RD Sharma XII Vol 2 2019 Solutions for Class 12 Commerce Math Chapter 9 Straight Line In Space are provided here with simple step-by-step explanations. These solutions for Straight Line In Space are extremely popular among class 12 Commerce students for Math Straight Line In Space Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma XII Vol 2 2019 Book of class 12 Commerce Math Chapter 9 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RD Sharma XII Vol 2 2019 Solutions. All RD Sharma XII Vol 2 2019 Solutions for class 12 Commerce Math are prepared by experts and are 100% accurate.

Page No 28.10:

Question 8:

Find the vector equation of a line passing through (2, −1, 1) and parallel to the line whose equations are x-32=y+17=z-2-3.

Answer:

We know that the vector equation of a line passing through a point with position vector a and parallel to the vector b is r = a + λ b.

Here,
a =2i^-j^+k^ b =2i^+7j^-3k^

Vector equation of the required line is
r = 2i^-j^+k^ + λ 2i^+7j^-3k^Here, λ is a parameter.

Page No 28.10:

Question 9:

The cartesian equations of a line are x-53=y+47=z-62. Find a vector equation for the line.

Answer:

The cartesian equation of the given line is x-53=y+47=z-62.
It can be re-written as

x-53=y--47=z-62

Thus, the given line passes through the point having position vector a=5i^-4j^+6k^ and is parallel to the vector b=3i^+7j^+2k^.

We know that the vector equation of a line passing through a point with position vector a and parallel to the vector b is r = a + λ b.

Vector equation of the required line is
r = 5i^-4j^+6k^ + λ 3i^+7j^+2k^Here, λ is a parameter.

Page No 28.10:

Question 10:

Find the cartesian equation of a line passing through (1, −1, 2) and parallel to the line whose equations are x-31=y-12=z+1-2. Also, reduce the equation obtained in vector form.

Answer:

We know that the cartesian equation of a line passing through a point with position vector a and parallel to the vector m is x-x1a=y-y2b=z-z3c.

Here,
a=x1i^+y1j^+z1k^ m=ai^+bj^+ck^

Here, a =i^-j^+2k^ and b =i^+2j^-2k^ 

Cartesian equation of the required line is

x-11=y--12=z-2-2x-11=y+12=z-2-2

We know that the cartesian equation of a line passing through a point with position vector a and parallel to the vector m is r=a+λm.
Here, the line is passing through the point 1, 1, -2 and its direction ratios are proportional to 1, 2, -2.

Vector equation of the required line is
r=i^-j^+2k^+λi^+2j^-2k^

Page No 28.10:

Question 11:

Find the direction cosines of the line 4-x2=y6=1-z3. Also, reduce it to vector form.

Answer:

The cartesian equation of the given line is

4-x2=y6=1-z3

It can be re-written as

x-4-2=y-06=z-1-3

This shows that the given line passes through the point 4, 0 ,1 and its direction ratios are proportional to -2, 6, -3.

So, its direction cosines are

-2-22+62+-32, 6-22+62+-32, -3-22+62+-32=-27, 67, -37 

Thus, the given line passes through the point having position vector a=4i^+k^ and is parallel to the vector b=-2i^+6j^-3k^.

We know that the vector equation of a line passing through a point with position vector a and parallel to the vector b is r = a + λ b.

Here,
a =4i^+k^ b = -2i^+6j^-3k^

Vector equation of the required line is
r = 4i^+0j^+k^ + λ -2i^+6j^-3k^                                 Here, λ is a parameter.

Page No 28.10:

Question 12:

The cartesian equations of a line are x = ay + b, z = cy + d. Find its direction ratios and reduce it to vector form.

Answer:

The cartesian equation of the given line is
x=ay+b, z=cy+d

It can be re-written as

x-ba=y-01=z-dc

Thus, the given line passes through the point b, 0, d and its direction ratios are proportional to a, 1, c. It is also parallel to the vector b=ai^+j^+ck^.

We know that the vector equation of a line passing through a point with position vector a and parallel to the vector b is r = a + λ b.

Vector equation of the required line is
r = bi^+0j^+dk^ + λ ai^+j^+ck^Here, λ is a parameter.

Page No 28.10:

Question 13:

Find the vector equation of a line passing through the point with position vector i^-2j^-3k^and parallel to the line joining the points with position vectors i^-j^+4k^ and 2i^+j^+2k^. Also, find the cartesian equivalent of this equation.

Answer:

We know that the vector equation of a line passing through a point with position vector a and parallel to the vector b is r = a + λ b.

Here,
a=i^-2j^-3k^ b= 2i^+j^+2k^-i^-j^+4k^=i^+2j^-2k^

Vector equation of the required line is
r = i^-2j^-3k^ + λ i^+2j^-2k^                                   ...(1)Here, λ is a parameter.

Reducing (1) to cartesian form, we get

xi^+yj^+zk^ = i^-2j^-3k^ + λ i^+2j^-2k^                      [Putting r= xi^+yj^+zk^ in (1)]xi^+yj^+zk^ = 1+λ i^+-2+2 λ j^+-3-2λ k^Comparing the coefficients of i^,  j^ and k^, we getx=1+λ, y=-2+2 λ, z=-3-2λx-11=λ, y+22=λ, z+3-2=λx-11=y+22=z+3-2=λHence, the cartesian form of (1) isx-11=y+22=z+3-2

Page No 28.10:

Question 14:

Find the points on the line x+23=y+12=z-32 at a distance of 5 units from the point P (1, 3, 3).

Answer:

The coordinates of any point on the line x+23=y+12=z-32 are given by

x+23=y+12=z-32=λx=3λ-2, y=2λ-1, z=2λ+3                          ...(1)

Let the coordinates of the desired point be 3λ-2, 2λ-1, 2λ+3 .

The distance between this point and (1, 3, 3) is 5 units.

 3λ-2-12+2λ-1-32+2λ+3-32=53λ-32+2λ-42+2λ2=2517λ2-34λ=0λλ-2=0λ=0 or 2

Substituting the values of λ in (1), we get the coordinates of the desired point as (-2, -1, 3) and (4, 3, 7).

Page No 28.10:

Question 15:

Show that the points whose position vectors are -2i^+3j^, i^+2j^+3k^ and 7i^-k^ are collinear.

Answer:

Let the given points be P, Q and R and let their position vectors be a, b and c, respectively. 

a=-2i^+3j^b=i^+2j^+3k^ c=7i^+9k^

Vector equation of line passing through P and Q is
r=a+λb-ar=-2i^+3j^+λi^+2j^+3k^--2i^+3j^r=-2i^+3j^+λ3i^-j^+3k^                         ...(1)

If points P, Q and R are collinear, then point R must satisfy (1).

 Replacing r by c=7i^+9k^ in (1), we get
7i^+9k^=-2i^+3j^+λ3i^-j^+3k^

Comparing the coefficients of i^, j^ and k^, we get
7=-2+3λ, 0=3-λ, 9=3λ
λ=3

These three equations are consistent, i.e. they give the same value of λ.
Hence, the given three points are collinear.

Disclaimer: The question given in the book has a minor error. The third vectors should be 7i^+9k^. The solution here is created accordingly.

 

Page No 28.10:

Question 16:

Find the cartesian and vector equations of a line which passes through the point (1, 2, 3) and is parallel to the line -x-21=y+37=2z-63.

Answer:

We have

-x-21=y+37=2z-63

It can be re-written as

x+2-1=y+37=z-332=x+2-2=y+314=z-33

This shows that the given line passes through the point -2, -3, 3 and its direction ratios are proportional to -2, 14, 3.

Thus, the parallel vector is b=-2i^+14j^+3k^.

We know that the vector equation of a line passing through a point with position vector a and parallel to the vector b is r = a + λ b.

Here,
a = i^+2j^+3k^ b=-2i^+14j^+3k^.

Vector equation of the required line is
r = i^+2j^+3k^ + λ -2i^+14j^+3k^                                ...(1)Here, λ is a parameter.

Reducing (1) to cartesian form, we get

xi^+yj^+zk^ = i^+2j^+3k^ + λ -2i^+14j^+3k^                      [Putting r=xi^+yj^+zk^ in (1)]xi^+yj^+zk^ = 1-2λ i^+2+14 λ j^+3+3λ k^Comparing the coefficients of i^,  j^ and k^, we getx=1-2λ, y=2+14 λ, z=3+3λx-1-2=λ, y-214=λ, z-33=λx-1-2=y-214=z-33=λHence, the cartesian form of (1) is x-1-2=y-214=z-33

Page No 28.10:

Question 17:

The cartesian equation of a line are 3x + 1 = 6y − 2 = 1 − z. Find the fixed point through which it passes, its direction ratios and also its vector equation.

Answer:

The cartesian equation of the given line is

3x + 1 = 6y − 2 = 1 − z

It can be re-written as

x+1313=y-1316=z-1-1=x--132=y-131=z-1-6

Thus, the given line passes through the point -13,13,1 and its direction ratios are proportional to 2, 1, -6. It is parallel to the vector b=2i^+j^-6k^.

We know that the vector equation of a line passing through a point with position vector a and parallel to the vector b is r = a + λ b.

Vector equation of the required line is
r = -13i^+13j^+k^ + λ 2i^+j^-6k^Here, λ is a parameter.

Page No 28.10:

Question 18:

Find the vector equation of the line passing through the point A(1, 2, –1) and parallel to the line 5x – 25 = 14 – 7y = 35z.

Answer:

The equation of the line 5x − 25 = 14 − 7y = 35z can be re-written as

      x-515=y-2-17=z135 x-57=y-2-5=z1

Since the required line is parallel to the given line, so the direction ratios of the required line are proportional to 7, -5, 1.

The vector equation of the required line passing through the point (1, 2, -1) and having direction ratios proportional to 7, -5, 1 is

r=i^+2j^-k^+λ7i^-5j^+k^.



Page No 28.15:

Question 1:

Show that the three lines with direction cosines 1213, -313, -413; 413, 1213, 313; 313, -413, 1213 are mutually perpendicular.

Answer:

The direction cosines of the three lines are

l1=1213, m1=-313, n1=-413l2=413, m2=1213, n1=313l3=313, m3=-413, n3=1213

 l1l2+m1m2+n1n2=48-36-12169=0Also,l2l3+m2m3+n2n3=12-48+36169=0l1l3+m1m3+n1n3=36+12-48169=0

Hence, the given lines are perpendicular to each other.

Page No 28.15:

Question 2:

Show that the line through the points (1, −1, 2) and (3, 4, −2) is perpendicular to the through the points (0, 3, 2) and (3, 5, 6).

Answer:

Suppose vector a is passing through the points (1, -1, 2) and (3, 4,-2) and b is passing through the points (0, 3, 2) and (3, 5, 6).

Then,
a=2i^+5j^-4k b=3i^+2j^+4k

Now,
a.b=2i^+5j^-4k.3i^+2j^+4k=0

Hence, the given lines are perpendicular to each other.



Page No 28.16:

Question 3:

Show that the line through the points (4, 7, 8) and (2, 3, 4) is parallel to the line through the points (−1, −2, 1) and, (1, 2, 5).

Answer:

Equations of lines passing through the points x1, y1, z1 and x2, y2, z2 are given by

x-x1x2-x1=y-y1y2-y1=z-z1z2-z1

So, the equation of a line passing through (4, 7, 8) and (2, 3, 4) is

x-42-4=y-73-7=z-84-8x-4-2=y-7-4=z-8-4

Also, the equation of the line passing through the points (-1, -2, 1) and (1, 2, 5) is

x+11+1=y+22+2=z-15-1x+12=y+24=z-14

We know that two lines are parallel if
a1a2=b1b2=c1c2 Cartesian equations of the two lines are given by x-x1a1=y-y1b1=z-z1c1 and x-x2a2=y-y2b2=z-z2c2


We observe

-22=-44=-44=-1

Hence, the given lines are parallel to each other.

Page No 28.16:

Question 4:

Find the cartesian equation of the line which passes through the point (−2, 4, −5) and parallel to the line given by x+33=y-45=z+86.

Answer:

We know that the cartesian equation of a line passing through a point with position vector a and parallel to the vector b is x-x1a=y-y2b=z-z3c.

Here,
a =-2i^+4j-5k^b =3i^+5j^-6k^ 

The cartesian equation of the required line is

x--23=y-45=z--56=x+23=y-45=z+56

Page No 28.16:

Question 5:

Show that the lines x-57=y+2-5=z1 and x1=y2=z3 are perpendicular to each other.

Answer:

We have

x-57=y+2-5=z1 and x1=y2=z3

These equations can be re-written as

x-57=y--2-5=z-01              ...(1) x-01=y-02=z-03                    ...(2)

 m1 =Vector parallel to line (1) =7i^-5j^+k^    m2 =Vector parallel to line (2) =i^+2j^+3k^  

Now,

m1.m2=7i^-5j^+k^.i^+2j^+3k^            =7-10+3            =0

Hence, the given two lines are perpendicular to each other.

Page No 28.16:

Question 6:

Show that the line joining the origin to the point (2, 1, 1) is perpendicular to the line determined by the points (3, 5, −1) and (4, 3, −1).

Answer:

The direction ratios of the line joining the origin to the point (2, 1, 1) are 2, 1, 1.

Let b1=2i^+j^+k^

The direction ratios of the line joining the points (3, 5, -1) and (4, 3, -1) are 1, -2, 0.

Let b2=i^-2j^+0k^

Now,

b1.b2=2i^+j^+k^.i^-2j^+0k^              =2-2+0              =0 b1b2

Hence, the two lines joining the given points are perpendicular to each other.

Page No 28.16:

Question 7:

Find the equation of a line parallel to x-axis and passing through the origin.

Answer:

The direction ratios of the line parallel to x-axis are proportional to 1, 0, 0.

Equation of the line passing through the origin and parallel to x-axis is

x-01=y-00=z-00=x1=y0=z0

Page No 28.16:

Question 8:

Find the angle between the following pairs of lines:
(i) r=4i^-j^+λi^+2j^-2k^ and r=i^-j^+2k^-μ2i^+4j^-4k^

(ii) r=3i^+2j^-4k^+λi^+2j^+2k^ and r=5j^-2k^+μ3i^+2j^+6k^

(iii) r=λi^+j^+2k^ and r=2j^+μ3-1i^-3+1j^+4k^

Answer:

(i) r=4i^-j^+λi^+2j^-2k^ and r=i^-j^+2k^-μ2i^+4j^-4k^

Let b1 and b2 be vectors parallel to the given lines.

Now,
b1=i^+2j^-2k^ b2=2i^+4j^-4k^

If θ is the angle between the given lines, then

cos θ=b1.b2b1 b2         =i^+2j^-2k^.2i^+4j^-4k^12+22+-22 22+42+-42         =2+8+83×6         =1θ=0°

(ii) r=3i^+2j^-4k^+λi^+2j^+2k^ and r=5j^-2k^+μ3i^+2j^+6k^

Let b1 and b2 be vectors parallel to the given lines.

Now,
b1=i^+2j^+2k^ b2=3i^+2j^+6k^

If θ is the angle between the given lines, then

cos θ=b1.b2b1 b2         =i^+2j^+2k^.3i^+2j^+6k^12+22+22 32+22+62         =3+4+123×7         =1921θ=cos-11921

(iii) r=λi^+j^+2k^ and r=2j^+μ3-1i^-3+1j^+4k^

Let b1 and b2 be vectors parallel to the given lines.

Now,
b1=i^+j^+2k^ b2=3-1i^-3+1j^+4k^

If θ is the angle between the given lines, then

cos θ=b1.b2b1 b2         =i^+j^+2k^.3-1i^-3+1j^+4k^12+12+22 3-12+3+12+42         =3-1-3+1+86 24         =612         =12θ=π3

Page No 28.16:

Question 9:

Find the angle between the following pairs of lines:
(i) x+43=y-15=z+34 and x+11=y-41=z-52

(ii) x-12=y-23=z-3-3 and x+3-1=y-58=z-14

(iii) 5-x-2=y+31=1-z3 and x3=1-y-2=z+5-1

(iv) x-23=y+3-2, z=5 and x+11=2y-33=z-52

(v) x-51=2y+6-2=z-31 and x-23=y+14=z-65

(vi) -x+2-2=y-17=z+3-3 and x+2-1=2y-84=z-54

Answer:

(i) x+43=y-15=z+34 and x+11=y-41=z-52

Let b1 and b2 be vectors parallel to the given lines.

b1=3i^+5j^+4k^ b2=i^+j^+2k^

If θ is the angle between the given lines, then

cos θ=b1.b2b1 b2         =3i^+5j^+4k^.i^+j^+2k^32+52+42 12+12+22         =3+5+8103         =853θ=cos-1853 


(ii) x-12=y-23=z-3-3 and x+3-1=y-58=z-14

Let b1 and b2 be vectors parallel to the given lines.

Now,
b1=2i^+3j^-3k^ b2=-i^+8j^+4k^

If θ is the angle between the given lines, then

cos θ=b1.b2b1 b2         =2i^+3j^-3k^.-i^+8j^+4k^22+32+-32 -12+82+42         =-2+24-12922         =10922θ=cos-110922 


(iii) 5-x-2=y+31=1-z3 and x3=1-y-2=z+5-1

The equations of the given lines can be re-written as

x-52=y+31=z-1-3 and x3=y-12=z+5-1

Let b1 and b2 be vectors parallel to the given lines.

Now,
b1=2i^+j^-3k^ b2=3i^+2j^-k^

If θ is the angle between the given lines, then

cos θ=b1.b2b1 b2         =2i^+j^-3k^.3i^+2j^-k^22+12+-32 32+22+-12         =6+2+314 14         =1114θ=cos-11114 


(iv) x-23=y+3-2, z=5 and x+11=2y-33=z-52

The equations of the given lines can be re-written as

x-23=y+3-2=z-50 and x+11=y-3232=z-52

Let b1 and b2 be vectors parallel to the given lines.

Now,
b1=3i^-2j^+0k^ b2=i^+32j^+2k^

If θ is the angle between the given lines, then

cos θ=b1.b2b1 b2         =3i^-2j^+0k^.i^+32j^+2k^32+-22+02 12+322+22         =3-3+013 294         =0θ=π2


(v ) x-51=2y+6-2=z-31 and x-23=y+14=z-65

The equations of the given lines can be re-written as

x-51=y+3-1=z-31 and x-23=y+14=z-65

Let b1 and b2 be vectors parallel to the given lines.

Now,
b1=i^-j^+k^b2=3i^+4j^+5k^

If θ is the angle between the given lines, then

cos θ=b1.b2b1 b2         =i^-j^+k^.3i^+4j^+5k^12+-12+12 32+42+52         =3-4+53 50         =456θ=cos-1456

Disclaimer: The answer given in the book is incorrect. This solution is created according to the question given in the book.


(vi)  -x+2-2=y-17=z+3-3 and x+2-1=2y-84=z-54

The equations of the given lines can be re-written as

x-22=y-17=z+3-3 and x+2-1=y-42=z-54

Let b1 and b2 be vectors parallel to the given lines.

Now,

b1=2i^+7j^-3k^ b2=-1i^+2j^+4k^

If θ is the angle between the given lines, then

cos θ=b1.b2b1 b2         =2i^+7j^-3k^.-1i^+2j^+4k^22+72+-32 -12+22+42         =-2+14-1262 21         =0θ=π2

Page No 28.16:

Question 10:

Find the angle between the pairs of lines with direction ratios proportional to
(i) 5, −12, 13 and −3, 4, 5
(ii) 2, 2, 1 and 4, 1, 8
(iii) 1, 2, −2 and −2, 2, 1
(iv) a, b, c and bc, ca, ab.

Answer:

(i) 5, −12, 13 and −3, 4, 5

Let m1 and m2 be vectors parallel to the two given lines. Then, the angle between the two given lines is same as the angle between m1  and  m2.Now, m1=Vector parallel to the line having direction ratios proportional to 5,-12, 13 m2=Vector parallel to the line having direction ratios proportional to -3, 4, 5 m1=5i^-12j^+13k^      m2=-3i^+4j^+5k^Let θ be the angle between the lines. 

Now,cos θ=m1.m2m1 m2         =5i^-12j^+13k^.-3i^+4j^+5k^52+-122+132 -32+42+52         =-15-48+65132×52         =165θ=cos-1165

(ii) 2, 2, 1 and 4, 1, 8

Let m1 and m2 be vectors parallel to the given two lines . Then, the angle between the lines is same as the angle between m1  and  m2.Now, m1=Vector parallel to the line having direction ratios proportional to  2, 2, 1m2=Vector parallel to the line having direction ratios proportional  to 4, 1, 8 m1=2i^-2j^+k^       m2=4i^+j^+8k^Let θ be the angle between the lines. 

Now,cos θ=m1.m2m1 m2         =2i^+2j^+k^.4i^+j^+8k^22+22+12 42+12+82         =8+2+83×9         =23θ=cos-123

(iii) 1, 2, −2 and −2, 2, 1

Let m1 and m2 be vectors parallel to the two given lines. Then, the angle between the two given lines is same as the angle between m1  and  m2.Now, m1=Vector parallel to the line having direction ratios proportional to 1, 2, -2 m2=Vector parallel to the line  having direction ratios proportional to-2, 2, 1 m1=i^+2j^-2k^       m2=-2i^+2j^+k^Let θ be the angle between the lines. 

Now,cos θ=m1.m2m1 m2         =i^+2j^-2k^.-2i^+2j^+k^12+22+-22 -22+22+12         =-2+4-23×3         =0θ=π2

(iv) a, b, c and bc, ca, ab.

Let m1 and m2 be vectors parallel to the given two lines. Then, the angle between the two lines is same as the angle between m1  and  m2.Now, m1=Vector parallel to the line having direction ratios proportional to a, b, cm2=Vector parallel to the line having direction ratios proportional to b-c, c-a, a-b m1=ai^+bj^+ck^ and m2=b-ci^+c-aj^+a-bk^Let θ be the angle between the lines.

Now,cos θ=m1.m2m1 m2         =ai^+bj^+ck^.b-ci^+c-aj^+a-bk^a2+b2+c2 b-c2+c-a2+a-b2         =ab-ac+bc-ba+ca-cba2+b2+c2 b-c2+c-a2+a-b2         =0θ=π2

Page No 28.16:

Question 11:

Find the angle between two lines, one of which has direction ratios 2, 2, 1 while the  other one is obtained by joining the points (3, 1, 4) and (7, 2, 12).

Answer:

The direction ratios of the line joining the points (3, 1, 4) and (7, 2, 12) are proportional to 4, 1, 8.

Let m1 and m2 be vectors parallel to the lines having direction ratios proportional to 2, 2, 1 and 4, 1, 8.

Now,
b1=2i^+2j^+k^ b2=4i^+j^+8k^

If θ is the angle between the given lines, then

cos θ=m1.m2m1 m2         =2i^+2j^+k^.4i^+j^+8k^22+22+12 42+12+82         =8+2+83×9         =23θ=cos-123

Page No 28.16:

Question 12:

Find the equation of the line passing through the point (1, 2, −4) and parallel to the line x-34=y-52=z+13.

Answer:

The direction ratios of the line parallel to line x-34=y-52=z+13 are proportional to 4, 2, 3.

Equation of the required line passing through the point (1, 2, -4) having direction ratios proportional to 4, 2, 3 is

x-14=y-22=z--43=x-14=y-22=z+43

Page No 28.16:

Question 13:

Find the equations of the line passing through the point (−1, 2, 1) and parallel to the line 2x-14=3y+52=2-z3.

Answer:

The equation of line 2x-14=3y+52=2-z3 can be re-written as

x-122=y+5323=z-2-3

The direction ratios of the line parallel to line 2x-14=3y+52=2-z3 are proportional to 2, 23, -3.

Equation of the required line passing through the point (-1, 2, 1) having direction ratios proportional to 2, 23, -3 is

x--12=y-223=z-1-3=x+12=y-223=z-1-3

Page No 28.16:

Question 14:

Find the equation of the line passing through the point (2, −1, 3) and parallel to the line r=i^-2j^+k^+λ2i^+3j^-5k^.

Answer:

The given line is parallel to the vector 2i^+3j^-5k^ and the required line is parallel to the given line.
So, the required line is parallel to the vector 2i^+3j^-5k^.

Hence, the equation of the required line passing through the point (2,-1, 3) and parallel to the vector 2i^+3j^-5k^ is r=2i^-j^+3k^+λ2i^+3j^-5k^.

Page No 28.16:

Question 15:

Find the equations of the line passing through the point (2, 1, 3) and perpendicular to the lines
x-11=y-22=z-33 and x-3=y2=z5

Answer:

Let:
b1=i^+2j^+3k^ b2=-3i^+2j^+5k^

Since the required line is perpendicular to the lines parallel to the vectors b1=i^+2j^+3k^ and b2=-3i^+2j^+5k^, it is parallel to the vector b=b1×b2.

Now,
b=b1×b2   =i^j^k^123-325   =4i^-14j^+8k^   =22i^-7j^+4k^

Thus, the direction ratios of the required line are proportional to 2, -7, 4.

The equation of the required line passing through the point (2, 1, 3) and having direction ratios proportional to 2, -7, 4 is x-22=y-1-7=z-34.



Page No 28.17:

Question 16:

Find the equation of the line passing through the point i^+j^-3k^ and perpendicular to the lines r=i^+λ2i^+j^-3k^ and r=2i^+j^-k^+μi^+j^+k^.

Answer:

The required line is perpendicular to the lines parallel to the vectors b1=2i^+j^-3k^ and b2=i^+j^+k^.

So, the required line is parallel to the vector b=b1×b2.

Now,
b=b1×b2   =i^j^k^21-3111   =4i^-5j^+k^

Equation of the required line passing through the point i^+j^-3k^ and parallel to 4i^-5j^+k^ is r=i^+j^-3k^+λ4i^-5j^+k^.

Page No 28.17:

Question 17:

Find the equation of the line passing through the point (1, −1, 1) and perpendicular to the lines joining the points (4, 3, 2), (1, −1, 0) and (1, 2, −1), (2, 1, 1).

Answer:

The direction ratios of the line joining the points (4, 3, 2), (1,-1, 0) and (1, 2, -1), (2, 1, 1) are -3, -4, -2 and 1, -1, 2, respectively.

Let:
b1=-3i^-4j^-2k^ b2=i^-j^+2k^

Since the required line is perpendicular to the lines parallel to the vectors b1=-3i^-4j^-2k^ and b2=i^-j^+2k^, it is parallel to the vector b=b1×b2.

Now,
b=b1×b2   =i^j^k^-3-4-21-12   =-10i^+4j^+7k^

So, the direction ratios of the required line are proportional to -10, 4, 7.

The equation of the required line passing through the point (1, -1, 1) and having direction ratios proportional to -10, 4, 7 is x-1-10=y+14=z-17.

Page No 28.17:

Question 18:

Determine the equations of the line passing through the point (1, 2, −4) and perpendicular to the two lines x-88=y+9-16=z-107 and x-153=y-298=z-5-5.

Answer:

We have

x-88=y+9-16=z-107x-153=y-298=z-5-5

Let:
b1=8i^-16j^+7k^ b2=3i^+8j^-5k^

Since the required line is perpendicular to the lines parallel to the vectors b1=8i^-16j^+7k^ and b2=3i^+8j^-5k^, it is parallel to the vector b=b1×b2.

Now,
b=b1×b2  =i^j^k^8-16738-5   =24i^+61j^+112k^

The direction ratios of the required line are proportional to 24, 61, 112.

The equation of the required line passing through the point (1, 2, -4) and having direction ratios proportional to 24, 61, 112 is x-124=y-261=z+4112.

Page No 28.17:

Question 19:

Show that the lines x-57=y+2-5=z1 and x1=y2=z3 are perpendicular to each other.

Answer:

The direction ratios of the lines x-57=y+2-5=z1 and x1=y2=z3 are proportional to 7, -5, 1 and 1, 2, 3, respectively.

Let:
b1=7i^-5j^+k^ b2=i^+2j^+3k^

Now,

b1.b2=7i^-5j^+k^.i^+2j^+3k^              =7-10+3              =0 b1b2

Hence, the given lines are perpendicular to each other.

Page No 28.17:

Question 20:

Find the vector equation of the line passing through the point (2, −1, −1) which is parallel to the line 6x − 2 = 3y + 1 = 2z − 2.

Answer:

The equation of the line 6x − 2 = 3y + 1 = 2z − 2 can be re-written as

x-1316=y+1313=z-112= x-131=y+132=z-13

Since the required line is parallel to the given line, the direction ratios of the required line are proportional to 1, 2, 3.

The vector equation of the required line passing through the point (2, -1, -1) and having direction ratios proportional to 1, 2, 3 is r=2i^-j^-k^+λi^+2j^+3k^.

Page No 28.17:

Question 21:

If the lines x-1-3=y-22 λ=z-32 and x-13λ=y-11=z-6-5 are perpendicular, find the value of λ.

Answer:

The equations of the given lines are

 x-1-3=y-22 λ=z-32x-13λ=y-11=z-6-5

Since the given lines are perpendicular to each other, we have

-33λ+2λ1+2-5=0-9λ+2λ-10=0λ=-107

Page No 28.17:

Question 22:

If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (−4, 3, −6) and (2, 9, 2) respectively, then find the angle between the lines AB and CD.

Answer:

The direction ratios of AB and CD are proportional to 3, 3, 4 and 6, 6, 8, respectively.

Let θ be the angle between AB and CD. Then,

cos θ=3×6+3×6+4×832+32+42 62+62+82         =6834 136         =1θ=0°

Page No 28.17:

Question 23:

Find the value of λ so that the following lines are perpendicular to each other. x-55λ+2=2-y5=1-z-1, x1=2y+14λ=1-z-3

Answer:

The equations of the given lines  x-55λ+2=2-y5=1-z-1 and x1=2y+14λ=1-z-3 can be re-written as

x-55λ+2=y-2-5=z-11 and x1=y+122λ=z-13


Since the given lines are perpendicular to each other, we have

5λ+21-52λ+13=05λ=5λ=1

Page No 28.17:

Question 24:

Find the direction cosines of the line x+22=2y-76=5-z6. Also, find the vector equation of the line through the point A(−1, 2, 3) and parallel to the given line.                                                              [CBSE 2014]

Answer:


The equation of the given line is x+22=2y-76=5-z6.
The given equation can be re-written as x+22=y-723=z-5-6.

This line passes through the point -2,72,5 and has direction ratios proportional to 2, 3, −6.

So, its direction cosines are

222+32+-62,322+32+-62,-622+32+-62

Or 27,37,-67

The required line passes through the point having position vector a=-i^+2j^+3k^ and is parallel to the vector b=2i^+3j^-6k^.

So, its vector equation is

r=-i^+2j^+3k^+λ2i^+3j^-6k^



Page No 28.22:

Question 1:

Show that the lines x1=y-22=z+33 and x-22=y-63=z-34 intersect and find their point of intersection.

Answer:

The coordinates of any point on the first line are given by

x1=y-22=z+33=λx=λ     y=2λ+2     z=3λ-3

The coordinates of a general point on the first line are λ, 2λ+2, 3λ-3.

Also, the coordinates of any point on the second line are given by

x-22=y-63=z-34=μx=2μ+2    y=3μ+6    z=4μ+3

The coordinates of a general point on the second line are 2μ+2, 3μ+6, 4μ+3

If the lines intersect, then they have a common point. So, for some values of λ and μ, we must have

λ=2μ+2, 2λ+2=3μ+6, 3λ-3=4μ+3λ-2μ=2                  ...(1)    2λ-3μ=4                ...(2)    3λ-4μ=6                ...(3)Solving (1) and (2), we getλ=2 and μ=0Substituting λ=2 and μ=0 in (3), we getLHS=3λ-4μ        =32-40        =6        =RHSSince λ=2 and μ=0 satisfy the third equation, the given lines intersect at 2, 6, 3.

Page No 28.22:

Question 2:

Show that the lines x-13=y+12=z-15 and x+24=y-13=z+1-2 do not intersect.

Answer:

The coordinates of any point on the first line are given by

x-13=y+12=z-15=λx=3λ+1     y=2λ-1     z=5λ+1

The coordinates of a general point on the first line are 3λ+1, 2λ-1, 5λ+1.

The coordinates of any point on the second line are given by

x+24=y-13=z+1-2=μx=4μ-2    y=3μ+1     z=-2μ-1

The coordinates of a general point on the second line are 4μ-2, 3μ+1, -2μ-1.

If the lines intersect, then they have a common point. So, for some values of λ and μ, we must have

3λ+1=4μ-2, 2λ-1=3μ+1, 5λ+1=-2μ-13λ-4μ=-3                 ...(1)    2λ-3μ=2                      ...(2) 5λ+2μ=-2                     ...(3)Solving (1) and (2), we getλ=-17 μ=-12Substituting λ=-17 and μ=-12 in (3), we getLHS=3λ+2μ       =3-17+2-12       =-75       -2LHSRHSHence, the given lines do not intersect.

Page No 28.22:

Question 3:

Show that the lines x+13=y+35=z+57 and x-21=y-43=z-65 intersect. Find their point of intersection.

Answer:

The coordinates of any point on the first line are given by

x+13=y+35=z+57=λx=3λ-1    y=5λ-3   z=7λ-5

The coordinates of a general point on the first line are 3λ-1, 5λ-3, 7λ-5.

The coordinates of any point on the second line are given by

x-21=y-43=z-65=μx=μ+2     y=3μ+4     z=5μ+6

The coordinates of a general point on the second line are μ+2, 3μ+4, 5μ+6.

If the lines intersect, then they have a common point. So, for some values of λ and μ, we must have

3λ-1=μ+2, 5λ-3=3μ+4, 7λ-5=5μ+63λ-μ=3                      ...(1)    5λ-3μ=7                     ...(2)    7λ-5μ=11                   ...(3)Solving (1) and (2), we getλ=12 μ=-32Substituting λ=12 and μ=-32 in (3), we getLHS=7λ-5μ       =712-5-32       =11        =RHSSince λ=12 and μ=-32 satisfy (3), the given lines intersect.Substituting the value of λ in the general coordinates of the first line, we getx=12y=-12z=-32Hence, the given lines intersect at point 12, -12, -32.

Page No 28.22:

Question 4:

Prove that the lines through A (0, −1, −1) and B (4, 5, 1) intersects the line through C (3, 9, 4) and D (−4, 4, 4). Also, find their point of intersection.

Answer:

The coordinates of any point on the line AB are given by

x-04-0=y+15+1=z+11+1=λx=4λ    y=6λ-1    z=2λ-1

The coordinates of a general point on AB are 4λ, 6λ-1, 2λ-1.

The coordinates of any point on the line CD are given by

x-33+4=y-99-4=z-44-4=μx=7μ+3    y=5μ+9    z=4

The coordinates of a general point on CD are 7μ+3, 5μ+9, 4.
If the lines AB and CD intersect, then they have a common point. So, for some values of λ and μ, we must have

4λ=7μ+3, 6λ-1=5μ+9, 2λ-1=44λ-7μ=3               ...(1)     6λ-5μ=10             ...(2)      λ=52                       ...(3)Solving (2) and (3), we getλ=52 μ=1Substituting λ=52 and μ=1 in (1), we getLHS=4λ-7μ        =452-71        =3       =RHSSince λ=52 and μ=1 satisfy (3), the given lines intersect.Substituting the value of λ in the coordinates of a general point on the line AB, we getx=10y=14 z=4Hence, AB and CD intersect at point 10, 14, 4.

Page No 28.22:

Question 5:

Prove that the line r=i^+j^-k^+λ3i^-j^ and r=4i^-k^+μ2i^+3k^ intersect and find their point of intersection.

Answer:

The position vectors of two arbitrary points on the given lines are
i^+j^-k^+λ3i^-j^=1+3λi^+ 1-λj^-k^4i^-k^+μ2i^+3k^=4+2μi^+0j^+3μ-1k^ 

If the lines intersect, then they have a common point. So, for some values of λ and μ, we must have

1+3λi^+ 1-λj^-k^=4+2μi^+0j^+3μ-1k^

Equating the coefficients of i^, j^ and k^, we get

1+3λ=4+2μ                ...(1)1-λ=0                          ...(2)3μ-1=-1                    ...(3)

Solving (2) and (3), we get
λ=1 μ=0

Substituting the values λ=1 and μ=0 in (1), we get

LHS=1+3λ       =1+31       =4RHS=4+2μ        =4+20        =4LHS=RHSSince λ=1 and μ=0 satisfy (3), the given lines intersect.

Substituting μ=0 in the second line, we get r=4i^+0j^-k^ as the position vector of the point of intersection.

Thus, the coordinates of the point of intersection are (4, 0, -1).

Page No 28.22:

Question 6:

Determine whether the following pair of lines intersect or not:
(i) r=i^-j^+λ2i^+k^ and r=2i^-j^+μi^+j^-k^

(ii) x-12=y+13=z and x+15=y-21; z=2

(iii) x-13=y-1-1=z+10 and x-42=y-00=z+13

(iv) x-54=y-74=z+3-5 and x-87=y-41=3-53

Answer:

(i) r=i^-j^+λ2i^+k^ and r=2i^-j^+μi^+j^-k^

The position vectors of two arbitrary points on the given lines are

i^-j^+λ2i^+k^=1+2λ i^ -j^+λk^2i^-j^+μi^+j^-k^=2+μ i^+-1+μ j^-μk^ 

If the lines intersect, then they have a common point. So, for some values of λ and μ, we must have

1+2λi^+ -j^+λk^=2+μi^+-1+μj^-μk^

Equating the coefficients of i^, j^ and k^, we get

1+2λ=2+μ             ...(1)-1=-1+μ              ...(2) λ=-μ                        ...(3)

Solving (2) and (3), we get
λ=0 μ=0.

Substituting the values λ=0 and μ=0 in (1), we get
LHS=1+2λ       =1+20       =1RHS=2+μ        =2+0        =2LHSRHSSince λ=0 and μ=0 do not satisfy (1), the given lines do not intersect.


(ii) x-12=y+13=z and x+15=y-21; z=2

The coordinates of any point on the first line are given by

x-12=y+13=z=λx=2λ+1     y=3λ-1      z=λ

The coordinates of a general point on the first line are 2λ+1, 3λ-1, λ.

Also, the coordinates of any point on the second line are given by

x+15=y-21=μ, z=2x=5μ-1     y=μ+2      z=2

The coordinates of a general point on the second line are 5μ-1, μ+2, 2.

If the lines intersect, then they have a common point. So, for some values of λ and μ, we must have

2λ+1=5μ-1, 3λ-1=μ+2, λ=2 2λ-5μ=-2                       ...(1)      3λ-μ=3                             ...(2)      λ=2                                     ...(3)Solving (2) and (3), we getλ=2 μ=3Substituting λ=2 and μ=3 in 1, we getLHS=2λ-5μ        =22-53        =4-15        =-11-2LHSRHSSince λ=2 and μ=3 do not satisfy (1), the given lines do not intersect.


(iii) x-13=y-1-1=z+10 and x-42=y-00=z+13

The coordinates of any point on the first line are given by

x-13=y-1-1=z+10=λx=3λ+1     y=-λ+1      z=-1

The coordinates of a general point on the first line are 3λ+1, -λ+1, -1.

Also, the coordinates of any point on the second line are given by

x-42=y-00=z+13=μx=2μ+4     y=0    z=3μ-1

The coordinates of a general point on the second line are 2μ+4, 0, 3μ-1.

If the lines intersect, then they have a common point. So, for some values of λ and μ, we must have

3λ+1=2μ+4, -λ+1=0, -1=3μ-13λ-2μ=3                ...(1)     λ=1                           ...(2)     μ=0                           ...(3)From (2) and (3), we getλ=1μ=0Substituting λ=1 and μ=0 in (1), we getLHS=3λ-2μ        =31-20        =3        =RHSSince λ=1 and μ=0 satisfy (1), the lines intersect.Substituting λ=1 and μ=0 in the coordinates of a general point on the first line, we getx=4y=0 z=-1Hence, the given lines intersect at 4, 0, -1.


(iv) x-54=y-74=z+3-5 and x-87=y-41=z-53

The coordinates of any point on the first line are given by

x-54=y-74=z+3-5=λx=4λ+5    y=4λ+7    z=-5λ-3

The coordinates of a general point on the first line are 4λ+5, 4λ+7, -5λ-3.

The coordinates of any point on the second line are given by

x-87=y-41=z-53=μx=7μ+8    y=μ+4    z=3μ+5

The coordinates of a general point on the second line are 7μ+8, μ+4, 3μ+5.

If the lines intersect, then they have a common point. So, for some values of λ and μ, we must have

4λ+5=7μ+8, 4λ+7=μ+4, -5λ-3=3μ+54λ-7μ=3             ...(1)    4λ-μ=-3            ...(2)    5λ+3μ=-8          ...(3)From (1) and (2), we getλ=-1 μ=-1Substituting λ=-1 and μ=-1 in (3), we getLHS=5λ+3μ         =5-1+3-1        =-8        =RHSSince λ=-1 and μ=-1 satisfy (3), the lines intersect.Substituting λ=-1 and μ=-1 in the coordinates of a general point on the first line, we getx=1y=3z=2Hence, the given lines intersect at 1, 3, 2.

Disclaimer: The question printed in the book is incorrect. Instead of z, 3 is printed.



Page No 28.23:

Question 7:

Show that the lines r=3i^+2j^-4k^+λi^+2j^+2k^ and r=5i^-2j^+μ3i^+2j^+6k^ are intersecting. Hence, find their point of intersection.

Answer:

The position vectors of two arbitrary points on the given lines are

3i^+2j^-4k^+λi^+2j^+2k^=3+λi^+ 2+2λj^+2λ-4k^5i^-2j^+μ3i^+2j^+6k^=5+3μi^+-2+2μj^+6μk^ 

If the lines intersect, then they have a common point. So, for some values of λ and μ, we must have
3+λi^+ 2+2λj^+2λ-4k^=5+3μi^+-2+2μj^+6μk^

Equating the coefficients of i^, j^ and k^, we get

3+λ=5+3μ               ...(1)2+2λ=-2+2μ         ...(2) 2λ-4=6μ                   ...(3)

Solving (1) and (2), we get
λ=-4μ=-2.

Substituting the values λ=-4 and μ=-2 in (3), we get

LHS=2λ-4       =2-4-4       =-12RHS=6μ        =6-2        =-12LHS=RHSSince λ=-4 and μ=-2 satisfy (3), the lines intersect.

Substituting μ=-2 in the second line, we get r=5i^-2j^-6i^-4j^-12k^=-i^-6j^-12k^ as the position vector of the point of intersection.

Thus, the coordinates of the point of intersection are (-1, -6, -12).



Page No 28.29:

Question 1:

Find the perpendicular distance of the point (3, −1, 11) from the line x2=y-2-3=z-34.

Answer:

Let the point (3, -1, 11) be P and the point through which the line passes be Q (0, 2, 3).
The line is parallel to the vector b=2i^-3j^+4k^.

Now,

PQ=-3i^+3j^-8k^
 b×PQ=i^j^k^2-34-33-8                 =12i^+4j^+15k^b×PQ=122+42+152                    =144+16+225                    =385d=b×PQb  =38529  =38529

Disclaimer: The answer given in the book is incorrect. This solution is created according to the question given in the book.

Page No 28.29:

Question 2:

Find the perpendicular distance of the point (1, 0, 0) from the line x-12=y+1-3=z+108. Also, find the coordinates of the foot of the perpendicular and the equation of the perpendicular.

Answer:

Let the point (1, 0, 0) be P and the point through which the line passes be Q (1, -1, -10).
The line is parallel to the vector b=2i^-3j^+8k^.

Now,

PQ=0i^-j^-10k^
 b×PQ=i^j^k^2-380-1-10                 =38i^+20j^-2k^b×PQ=382+202+22                   =1444+400+4                   =1848d=b×PQb  =184877  =24  =26


Let L be the foot of the perpendicular drawn from the point P (1, 0, 0) to the given line.



The coordinates of a general point on the line x-12=y+1-3=z+108 are given by

x-12=y+1-3=z+108=λx=2λ+1     y=-3λ-1     z=8λ-10

Let the coordinates of L be 2λ+1, -3λ-1, 8λ-10.

The direction ratios of PL are proportional to 2λ+1-1, -3λ-1-0, 8λ-10-0, i.e. 2λ, -3λ-1, 8λ-10.

The direction ratios of the given line are proportional to 2 ,-3, 8, but PL is perpendicular to the given line.

 22λ-3-3λ-1+88λ-10=0λ=1

Substituting λ=1 in 2λ+1, -3λ-1, 8λ-10, we get the coordinates of L as (3, -4, -2).

Equation of the line PL is given by

x-13-1=y-0-4-0=z-0-2-0=x-11=y-2=z-1

r=i^+λi^-2j^-k^

Page No 28.29:

Question 3:

Find the foot of the perpendicular drawn from the point A (1, 0, 3) to the joint of the points B (4, 7, 1) and C (3, 5, 3).

Answer:

Let D be the foot of the perpendicular drawn from the point A (1, 0, 3) to the line BC.

The coordinates of a general point on the line BC are given by

x-44-3=y-77-5=z-11-3=λx=λ+4    y=2λ+7    z=-2λ+1

Let the coordinates of D be λ+4, 2λ+7, -2λ+1.




The direction ratios of AD are proportional to λ+4-1, 2λ+7-0, -2λ+1-3, i.e. λ+3, 2λ+7, -2λ-2.

The direction ratios of the line BC are proportional to 1, 2, -2, but AD is perpendicular to the line BC.

1λ+3+22λ+7-2-2λ-2=0λ=-73
Substituting λ=-73 in λ+4, 2λ+7, -2λ+1, we get the coordinates of D as 53, 73, 173.

Page No 28.29:

Question 4:

A (1, 0, 4), B (0, −11, 3), C (2, −3, 1) are three points and D is the foot of perpendicular from A on BC. Find the coordinates of D.

Answer:

Point D is the foot of the perpendicular drawn from the point A (1, 0, 4) to the line BC.

The coordinates of a general point on the line BC are given by

x-02-0=y+11-3+11=z-31-3=λx=2λ     y=8λ-11     z=-2λ+3

Let the coordinates of D be 2λ, 8λ-11, -2λ+3.

The direction ratios of AD are proportional to 2λ-1, 8λ-11-0, -2λ+3-4, i.e. 2λ-1, 8λ-11, -2λ-1.

The direction ratios of the line BC are proportional to 2, 8,-2, but AD is perpendicular to the line BC.

 22λ-1+88λ-11-2-2λ-1=0λ=119

Substituting λ=119 in 2λ, 8λ-11, -2λ+3, we get the coordinates of D as 229, -119, 59.

Page No 28.29:

Question 5:

Find the foot of perpendicular from the point (2, 3, 4) to the line 4-x2=y6=1-z3. Also, find the perpendicular distance from the given point to the line.

Answer:

Let L be the foot of the perpendicular drawn from the point P (2, 3, 4) to the given line.

The coordinates of a general point on the line 4-x2=y6=1-z3 are given by
4-x2=y6=1-z3=λThey can be re-written as x-4-2=y6=z-1-3=λx=-2λ+4     y=6λ    z=-3λ+1

Let the coordinates of L be -2λ+4, 6λ, -3λ+1.



The direction ratios of PL are proportional to -2λ+4-2, 6λ-3, -3λ+1-4, i.e. -2λ+2, 6λ-3, -3λ-3.

The direction ratios of the given line are proportional to -2, 6, -3, but PL is perpendicular to the given line.

-2-2λ+2+66λ-3-3-3λ-3=0λ=1349

Substituting λ=1349 in -2λ+4, 6λ, -3λ+1, we get the coordinates of L as  17049, 7849, 1049.

 PL=17049-22+7849-32+1049-42         =445412401         =90949         =37101 .


Hence, the length of the perpendicular from P on PL is 37101 units.



Page No 28.30:

Question 6:

Find the equation of the perpendicular drawn from the point P (2, 4, −1) to the line x+51=y+34=z-6-9. Also, write down the coordinates of the foot of the perpendicular from P.

Answer:

Let L be the foot of the perpendicular drawn from the point P (2, 4, -1) to the given line.

The coordinates of a general point on the line x+51=y+34=z-6-9 are given by
x+51=y+34=z-6-9=λx=λ-5     y=4λ-3    z=-9λ+6

Let the coordinates of L be λ-5, 4λ-3, -9λ+6.



The direction ratios of PL are proportional to λ-5-2, 4λ-3-4, -9λ+6+1, i.e. λ-7, 4λ-7, -9λ+7.

The direction ratios of the given line are proportional to 1, 4, -9, but PL is perpendicular to the given line.

1λ-7+44λ-7-9-9λ+7=0λ=1

Substituting λ=1 in λ-5, 4λ-3, -9λ+6, we get the coordinates of L as -4, 1, -3.

Equation of the line PL is

x-2-4-2=y-41-4=z+1-3+1=x-2-6=y-4-3=z+1-2

Page No 28.30:

Question 7:

Find the length of the perpendicular drawn from the point (5, 4, −1) to the line r=i^+λ2i^+9j^+5k^.

Answer:

Let the point (5, 4, -1) be P and the the point through which the line passes be Q (1, 0, 0).
The line is parallel to the vector b=2i^+9j^+5k^.

Now,

PQ=-4i^-4j^+k^
 b×PQ=i^j^k^295-4-41                 =29i^-22j^+28k^b×PQ=292+-222+282                   =841+484+784                   =2109b=22+92+52    =4+81+25    =110d=b×PQb   =2109110   =2109110

Page No 28.30:

Question 8:

Find the foot of the perpendicular drawn from the point i^+6j^+3k^ to the line r=j^+2k^+λi^+2j^+3k^. Also, find the length of the perpendicular

Answer:

Let L be the foot of the perpendicular drawn from the point P (i^+6j^+3k^) to the line r=j^+2k^+λi^+2j^+3k^.

Let the position vector L be
r=j^+2k^+λi^+2j^+3k^=λi^+1+2λj^+2+3λk^        ...(1)


Now,
PL=Position vector of L- Position vector of PPL=λi^+1+2λj^+2+3λk^-i^+6j^+3k^PL=λ-1i^+2λ-5j^+3λ-1k^                      ...(2)


Since PL is perpendicular to the given line, which is parallel to b=i^+2j^+3k^, we have

 PL.b=0λ-1i^+2λ-5j^+3λ-1k^.i^+2j^+3k^=0 1λ-1+22λ-5+33λ-1=0λ=1

Substituting λ=1 in (1), we get the position vector of L as i^+3j^+5k^.

Substituting λ=1 in (2), we get
PL=-3j^+2k^
      =-32+22      =13

Hence, the length of the perpendicular from point P on PL is 13 units.

Page No 28.30:

Question 9:

Find the equation of the perpendicular drawn from the point P (−1, 3, 2) to the line r=2j^+3k^+λ2i^+j^+3k^. Also, find the coordinates of the foot of the perpendicular from P.

Answer:

Let L be the foot of the perpendicular drawn from the point P (-1, 3, 2) to the line r=2j^+3k^+λ2i^+j^+3k^.



Let the position vector L be
r=2j^+3k^+λ2i^+j^+3k^=2λi^+2+λj^+3+3λk^        ...(1)

Now,

PL=Position vector of L- Position vector of PPL=2λi^+2+λj^+3+3λk^--i^+3j^+2k^PL=2λ+1i^+λ-1j^+3λ+1k^                              ...(2) 

Since PL is perpendicular to the given line, which is parallel to b=2i^+j^+3k^, we have

 PL.b=02λ+1i^+λ-1j^+3λ+1k^.2i^+j^+3k^=0 22λ+1+1λ-1+33λ+1=0λ=-27

Substituting λ=-27 in (1), we get the position vector of L as -47i^+127j^+157k^.

So, the coordinates of the foot of the perpendicular from P to the given line is L -47, 127, 157.
Substituting λ=-27 in (2), we get
PL=37i^-97j^+17k^

Equation of the perpendicular drawn from P to the given line is
r=Position vector of P +λPL   =-i^+3j^+2k^ +λ3i^-9j^+k^

Page No 28.30:

Question 10:

Find the foot of the perpendicular from (0, 2, 7) on the line x+2-1=y-13=z-3-2.

Answer:

Let L be the foot of the perpendicular drawn from the point P (0, 2, 7) to the given line.

The coordinates of a general point on the line x+2-1=y-13=z-3-2 are given by
x+2-1=y-13=z-3-2=λx=-λ-2    y=3λ+1    z=-2λ+3

Let the coordinates of L be -λ-2, 3λ+1, -2λ+3.



The direction ratios of PL are proportional to -λ-2-0, 3λ+1-2, -2λ+3-7, i.e. -λ-2, 3λ-1, -2λ-4.

The direction ratios of the given line are proportional to -1, 3, -2, but PL is perpendicular to the given line.

-1-λ-2+33λ-1-2-2λ-4=0λ=-12
Substituting λ=-12 in -λ-2, 3λ+1, -2λ+3, we get the coordinates of L as -32, -12, 4.

Disclaimer: The answer given in the book is incorrect. This solution is created according to the question given in the book.

Page No 28.30:

Question 11:

Find the foot of the perpendicular from (1, 2, −3) to the line x+12=y-3-2=z-1.

Answer:

Let L be the foot of the perpendicular drawn from the point P (1, 2, -3) to the given line.

The coordinates of a general point on the line x+12=y-3-2=z-1 are given by
x+12=y-3-2=z-1=λx=2λ-1     y=-2λ+3     z=-λ

Let the coordinates of L be 2λ-1, -2λ+3 , -λ.



The direction ratios of PL are proportional to 2λ-1-1, -2λ+3-2, -λ+3, i.e. 2λ-2, -2λ+1, -λ+3.

The direction ratios of the given line are proportional to 2, -2, -1, but PL is perpendicular to the given line.

 22λ-2-2-2λ+1-1-λ+3=0λ=1

Substituting λ=1 in 2λ-1, -2λ+3 , -λ, we get the coordinates of L as 1, 1, -1.

Page No 28.30:

Question 12:

Find the equation of line passing through the points A (0, 6, −9) and B (−3, −6, 3). If D is the foot of perpendicular drawn from a point C (7, 4, −1) on the line AB, then find the coordinates of the point D and the equation of line CD.

Answer:



Equation of line AB passing through the points A(0, 6,-9) and B(-3, -6, 3) is

x-0-3-0=y-6-6-6=z+93+9=x1=y-64=z+9-4

Here, D is the foot of the perpendicular drawn from C (7, 4, -1) on AB.

The coordinates of a general point on AB are given by

x1=y-64=z+9-4=λx=λ     y=4λ+6     z=-4λ-9

Let the coordinates of D be λ, 4λ+6, -4λ-9.
The direction ratios of CD are proportional to λ-7, 4λ+6-4, -4λ-9+1, i.e. λ-7, 4λ+2, -4λ-8.

The direction ratios of AB are proportional to 1, 4, -4, but CD is perpendicular to AB.

Substituting λ=-1 in λ, 4λ+6, -4λ-9, we get the coordinates of D as -1, 2, -5.

Equation of CD is
x-7-1-7=y-42-4=z+1-5+1=x-74=y-41=z+12

Page No 28.30:

Question 13:

Find the distance of the point (2, 4, −1) from the line x+51=y+34=z-6-9.                  [NCERT EXEMPLAR]

Answer:


We know that the distance d from point P to the line l having equation r=a+λb is given by d=b×PQb, where Q is any point on the line l.

The equation of the given line is x+51=y+34=z-6-9.

Let P(2, 4, −1) be the given point. Now, in order to find the required distance we need to find a point Q on the given line.

The given line passes through the point (−5, −3, 6). So, take this point as the required point Q(−5, −3, 6).

Also, the line is parallel to the vector b=i^+4j^-9k^.

Now, PQ=-5i^-3j^+6k^-2i^+4j^-k^=-7i^-7j^+7k^

b×PQ=i^j^k^14-9-7-77=-35i^+56j^+21k^b×PQ=-352+562+212=1225+3136+441=4802=492

Let d be the required distance.

d=b×PQb=4921+16+81=49298=49272=7

Thus, the distance of the given point from the given line is 7 units.

Page No 28.30:

Question 14:

Find the coordinates of the foot of perpendicular drawn from the point A(1, 8, 4) to the line joining the points B(0, −1, 3) and C(2, −3, −1).                 [NCERT EXEMPLAR]

Answer:


The Cartesian equation of the line joining points B(0, −1, 3) and C(2, −3, −1) is

x-02-0=y--1-3--1=z-3-1-3Or x2=y+1-2=z-3-4

Let L be the foot of the perpendicular drawn from the point A(1, 8, 4) to the line x2=y+1-2=z-3-4.

The coordinates of general point on the line x2=y+1-2=z-3-4 are given by

x2=y+1-2=z-3-4=λOr x=2λ,y=-2λ-1,z=-4λ+3

Let the coordinates of L be 2λ,-2λ-1,-4λ+3. Therefore, the direction ratios of AL are proportional to

2λ-1,-2λ-1-8,-4λ+3-4 or 2λ-1,-2λ-9,-4λ-1

Direction ratios of the given line are proportional to 2, −2, −4.

But, AL is perpendicular to the given line.

2×2λ-1+-2×-2λ-9+-4×-4λ-1=04λ-2+4λ+18+16λ+4=024λ+20=0λ=-56

Putting λ=-56 in 2λ,-2λ-1,-4λ+3, we get

2×-56,-2×-56-1,-4×-56+3=-53,23,193

Thus, the required coordinates of the foot of the perpendicular are -53,23,193.



Page No 28.37:

Question 1:

Find the shortest distance between the following pairs of lines whose vector equations are:

(i) r=3i^+8j^+3k^+λ3i^-j^+k^ and r=-3i^-7j^+6k^+μ-3i^+2j^+4k^

(ii) r=3i^+5j^+7k^+λi^-2j^+7k^ and r=-i^-j^-k^+μ7i^-6j^+k^

(iii) r=i^+2j^+3k^+λ2i^+3j^+4k^ and r=2i^+4j^+5k^+μ3i^+4j^+5k^

(iv) r=1-t i^+t-2 j^+3-t k^ and r=s+1 i^+2s-1 j^-2s+1 k^

(v) r=λ-1 i^+λ+1 j^-1+λ k^ and r=1-μ i^+2μ-1 j^+μ+2 k^

(vi) r=2i^-j^-k^+λ2i^-5j^+2k^ and, r=i^+2j^+k^+μi^-j^+k^

(vii) r=i^+j^+λ2i^-j^+k^ and, r=2i^+j^-k^+μ3i^-5j^+2k^

(viii) r=8+3λi^-9+16λj^+10+7λk^ and r=15i^+29j^+5k^+μ3i^+8j^-5k^              [NCERT EXEMPLAR]

Answer:

(i) r=3i^+8j^+3k^+λ3i^-j^+k^ and r=-3i^-7j^+6k^+μ-3i^+2j^+4k^

Comparing the given equations with the equations r=a1+λb1 and r=a2+μb2, we get

a1=3i^+8j^+3k^a2=-3i^-7j^+6k^b1=3i^-j^+k^ b2=-3i^+2j^+4k^ a2- a1=-6i^-15j^+3k^and b1×b2=i^j^k^3-11-324                    =-6i^-15j^+3k^ b1×b2=-62+-152+32                   =36+225+9                   =270a2- a1.b1×b2=-6i^-15j^+3k^.-6i^-15j^+3k^                                =36+225+9                                =270

The shortest distance between the lines r=a1+λb1 and r=a2+μb2 is given by

d=a2- a1.b1×b2 b1×b2  =270270  =270


(ii) r=3i^+5j^+7k^+λi^-2j^+7k^ and r=-i^-j^-k^+μ7i^-6j^+k^

Comparing the given equations with the equations r=a1+λb1 and r=a2+μb2, we get

a1=3i^+5j^+7k^a2=-i^-j^-k^b1=i^-2j^+7k^b2=7i^-6j^+k^ a2- a1=-4i^-6j^-8k^and b1×b2=i^j^k^1-277-61                    =40i^+48j^+8k^ b1×b2=402+482+82                    =1600+2304+64                    =3968a2- a1.b1×b2=-4i^-6j^-8k^.40i^+48j^+8k^                                =-160-288-64                                =-512

The shortest distance between the lines r=a1+λb1 and r=a2+μb2 is given by

d=a2- a1.b1×b2 b1×b2  =-5123968   =5123968

(iii) r=i^+2j^+3k^+λ2i^+3j^+4k^ and r=2i^+4j^+5k^+μ3i^+4j^+5k^

Comparing the given equations with the equationsr=a1+λb1 and r=a2+μb2, we get

a1=i^+2j^+3k^a2=2i^+4j^+5k^b1=2i^+3j^+4k^ b2=3i^+4j^+5k^ a2- a1=i^+2j^+2k^and b1×b2=i^j^k^234345                    =-i^+2j^-k^ b1×b2=-12+22+-12                    =1+4+1                    =6a2- a1.b1×b2=i^+2j^+2k^.-i^+2j^-k^                                =-1+4-2                                =1

The shortest distance between the lines r=a1+λb1 and r=a2+μb2 is given by

d=a2- a1.b1×b2 b1×b2  =16  =16

(iv) r=1-t i^+t-2 j^+3-t k^ and r=s+1 i^+2s-1 j^-2s+1 k^

The vector equations of the given lines can be re-written as

r=i^-2j^+3k^+t-i^+j^-k^ and r=i^-j^-k^+si^+2j^-2k^

Comparing the given equations with the equations r=a1+λb1 and r=a2+μb2, we get

a1=i^-2j^+3k^a2=i^-j^-k^b1=-i^+j^-k^ b2=i^+2j^-2k^ a2- a1=j^-4k^and b1×b2=i^j^k^-11-112-2                    =-3j^-3k^ b1×b2=-32+-32                    =9+9                    =32a2- a1.b1×b2=j^-4k^.-3j^-3k^                                =-3+12                                =9

The shortest distance between the line r=a1+λb1 and r=a2+μb2 is given by
d=a2- a1.b1×b2 b1×b2  =932  =32

(v) r=λ-1 i^+λ+1 j^-1+λ k^ and r=1-μ i^+2μ-1 j^+μ+2 k^

The vector equations of the given lines can be re-written as

r=-i^+j^-k^+λi^+j^-k^ and r=i^-j^+2k^+μ-i^+2j^+k^
Comparing the given equations with the equations r=a1+λb1 and r=a2+μb2, we get

a1=-i^+j^-k^a2=i^-j^+2k^b1=i^+j^-k^b2=-i^+2j^+k^ a2- a1=2i^-2j^+3k^and b1×b2=i^j^k^11-1-121                    =3i^+3k^ b1×b2=32+32                    =9+9                    =32a2- a1.b1×b2=2i^-2j^+3k^.3i^+3k^                                =6+9                                =15

The shortest distance between the lines r=a1+λb1 and r=a2+μb2 is given by

d=a2- a1.b1×b2 b1×b2  =1532  =52

(vi) r=2i^-j^-k^+λ2i^-5j^+2k^ and, r=i^+2j^+k^+μi^-j^+k^

Comparing the given equations with the equations r=a1+λb1 and r=a2+μb2, we get

a1=2i^-j^-k^a2=i^+2j^+k^b1=2i^-5j^+2k^ b2=i^-j^+k^ a2- a1=-i^+3j^+2k^and b1×b2=i^j^k^2-521-11                    =-3i^+3k^ b1×b2=-32+32                    =9+9                    =32a2- a1.b1×b2=-i^+3j^+2k^.-3i^+3k^                                =3+6                                =9

The shortest distance between the lines r=a1+λb1 and r=a2+μb2 is given by

d=a2- a1.b1×b2 b1×b2  =932  =32

(vii) r=i^+j^+λ2i^-j^+k^ and, r=2i^+j^-k^+μ3i^-5j^+2k^

Comparing the given equations with the equationsr=a1+λb1 and r=a2+μb2, we get

a1=i^+j^a2=2i^+j^-k^b1=2i^-j^+k^b2=3i^-5j^+2k^ a2- a1=i^-k^and b1×b2=i^j^k^2-113-52                    =3i^-j^-7k^ b1×b2=32+-12+-72                    =9+1+49                    =59a2- a1.b1×b2=i^-k^.3i^-j^-7k^                                =3+7                                =10

The shortest distance between the lines r=a1+λb1 and r=a2+μb2 is given by

d=a2- a1.b1×b2 b1×b2  =1059  =1059

(viii) The vector equations of the given lines can be re-written as

r=8i^-9j^+10k^+λ3i^-16j^+7k^ and r=15i^+29j^+5k^+μ3i^+8j^-5k^

Comparing the given equations with the equations r=a1+λb1 and r=a2+μb2, we get

a1=8i^-9j^+10k^

b1=3i^-16j^+7k^

a2=15i^+29j^+5k^

b2=3i^+8j^-5k^

a2-a1=15i^+29j^+5k^-8i^-9j^+10k^=7i^+38j^-5k^

b1×b2=i^j^k^3-16738-5=24i^+36j^+72k^b1×b2=242+362+722=576+1296+5184=7056=84

Also,

a2-a1.b1×b2=7i^+38j^-5k^.24i^+36j^+72k^=7×24+38×36+-5×72=168+1368-360=1176

We know that the shortest distance between the lines r=a1+λb1 and r=a2+μb2 is given by d=a2- a1.b1×b2 b1×b2.

∴ Required shortest distance between the given pairs of lines,

d=a2- a1.b1×b2 b1×b2  =117684  =14



Page No 28.38:

Question 2:

Find the shortest distance between the following pairs of lines whose cartesian equations are:
(i) x-12=y-23=z-34 and x-23=y-34=z-55

(ii) x-12=y+13=z and x+13=y-21; z=2

(iii) x-1-1=y+21=z-3-2 and x-11=y+12=z+1-2

(iv) x-31=y-5-2=z-71 and x+17=y+1-6=z+11

Answer:

(i) The equations of the given lines are

x-12=y-23=z-34      ...(1) x-23=y-34=z-55      ...(2) 

Since line (1) passes through the point (1, 2, 3) and has direction ratios proportional to 2, 3, 4, its vector equation is
r=a1+λb1                                                                   Here,a1=i^+2j^+3k^ b1=2i^+3j^+4k^

Also, line (2) passes through the point (2, 3, 5) and has direction ratios proportional to 3, 4, 5.
Its vector equation is
r=a2+μb2                                                                Here,a2=2i^+3j^+5k^ b2=3i^+4j^+5k^

Now,
 a2- a1=i^+j^+2k^and b1×b2=i^j^k^234345                    =-i^+2j^-k^ b1×b2=-12+22+-12                    =1+4+1                    =6and a2- a1.b1×b2=i^+j^+2k^.-i^+2j^-k^                                       =-1+2-2                                       =-1

The shortest distance between the lines r=a1+λb1 and r=a2+μb2 is given by

d=a2- a1.b1×b2 b1×b2  =-16  =16

(ii) The equations of the given lines are

x-12=y+13=z-01         ...(1) x+13=y-21=z-20        ...(2)

Since line (1) passes through the point (1, -1, 0) and has direction ratios proportional to 2, 3, 1, its vector equation is
r=a1+λb1                                                              Here,a1=i^-j^+0k^ b1=2i^+3j^+k^

Also, line (2) passes through the point (-1, 2, 2) and has direction ratios proportional to 3, 1, 0.
Its vector equation is
r=a2+μb2                                                                 Here,a2=-i^+2j^+2k^ b2=3i^+j^+0k^

Now,
 a2- a1=-2i^+3j^+2k^and b1×b2=i^j^k^231310                    =-i^+3j^-7k^ b1×b2=-12+32+-72                    =1+9+49                    =59and a2- a1.b1×b2=-2i^+3j^+2k^.-i^+3j^-7k^                                       =2+9-14                                       =-3

The shortest distance between the lines r=a1+λb1 and r=a2+μb2 is given by

d=a2- a1.b1×b2 b1×b2  =-359  =359

Disclaimer: The answer given in the book is incorrect. This solution is created according to the question given in the book.

(iii) x-1-1=y+21=z-3-2          ...(1)  x-11=y+12=z+1-2         ...(2) 

Since line (1) passes through the point (1, -2, 3) and has direction ratios proportional to -1, 1, -2, its vector equation is
r=a1+λb1                                                                  Here,a1=i^-2j^+3k^ b1=-i^+j^-2k^

Also, line (2) passes through the point (1, -1, -1) and has direction ratios proportional to 1, 2, -2.
Its vector equation is
r=a2+μb2                                                                   Here,a2=i^-j^-k^b2=i^+2j^-2k^

Now,
 a2- a1=j^-4k^and b1×b2=i^j^k^-11-212-2                    =2i^-4j^-3k^ b1×b2=22+-42+-32                    =4+16+9                    =29a2- a1.b1×b2=j^-4k^.2i^-4j^-3k^                                =-4+12                                =8

The shortest distance between the lines r=a1+λb1 and r=a2+μb2 is given by
d=a2- a1.b1×b2 b1×b2  =829 =829


(iv) x-31=y-5-2=z-71           ...(1) x+17=y+1-6=z+11          ...(2)

Since line (1) passes through the point (3, 5, 7) and has direction ratios proportional to 1, -2, 1, its vector equation is
r=a1+λb1                                                                 Here,a1=3i^+5j^+7k^ b1=i^-2j^+k^

Also, line (2) passes through the point (-1, -1, -1) and has direction ratios proportional to 7, -6, 1.
Its vector equation is
r=a2+μb2                                                              Here,a2=-i^-j^-k^ b2=7i^-6j^+k^

Now,
 a2- a1=-4i^-6j^-8k^and b1×b2=i^j^k^1-217-61                   =4i^+6j^+8k^ b1×b2=42+62+82                    =16+36+64                    =116a2- a1.b1×b2=-4i^-6j^-8k^.4i^+6j^+8k^                                =-16-36-64                                =-116

The shortest distance between the lines r=a1+λb1 and r=a2+μb2 is given by
d=a2- a1.b1×b2 b1×b2  =-116116  =116  =229

Page No 28.38:

Question 3:

By computing the shortest distance determine whether the following pairs of lines intersect or not:
(i) r=i^-j^+λ2i^+k^ and r=2i^-j^+μi^+j^-k^

(ii) r=i^+j^-k^+λ3i^-j^ and r=4i^-k^+μ2i^+3k^

(iii) x-12=y+13=z and x+15=y-21; z=2

(iv) x-54=y-7-5=z+3-5 and x-87=y-71=z-53

Answer:

(i) r=i^-j^+λ2i^+k^ and r=2i^-j^+μi^+j^-k^

Comparing the given equations with the equations r=a1+λb1 and r=a2+μb2, we get

a1=i^-j^a2=2i^-j^b1=2i^+k^ b2=i^+j^-k^ a2- a1=i^and b1×b2=i^j^k^20111-1                    =-i^+3j^+2k^a2- a1.b1×b2=i^.-i^+3j^+2k^                                =-1We observea2- a1.b1×b20Thus, the given lines do not intersect.          


(ii) r=i^+j^-k^+λ3i^-j^ and r=4i^-k^+μ2i^+3k^

Comparing the given equations with the equations r=a1+λb1 and r=a2+μb2, we get

a1=i^+j^-k^a2=4i^-k^b1=3i^-j^ b2=2i^+3k^ a2- a1=3i^-j^and b1×b2=i^j^k^3-10203                  =-3i^-9j^+2k^a2- a1.b1×b2=3i^-j^.-3i^-9j^+2k^                                =-9+9                                =0We observea2- a1.b1×b2=0Thus, the given lines intersect.


(iii) x-12=y+13=z-01 and x+15=y-21=z-20

Since the first line passes through the point (1, -1, 0) and has direction ratios proportional to 2, 3, 1, its vector equation is
r=a1+λb1                                                      ...(1)             Here,a1=i^-j^+0k^ b1=2i^+3j^+k^

Also, the second line passes through the point (-1, 2, 2) and has direction ratios proportional to 5, 1, 0.
Its vector equation is
r=a2+μb2                                                      ...(2)               Here, a2=-i^+2j^+2k^ b2=5i^+j^+0k^

Now,
 a2- a1=-2i^+3j^+2k^and b1×b2=i^j^k^231510                    =-i^+5j^-13k^a2- a1.b1×b2=-2i^+3j^+2k^.-i^+5j^-13k^                                =2+15-26                                =-9We observea2- a1.b1×b20Thus, the given lines do not intersect.


(iv) x-54=y-7-5=z+3-5 and x-87=y-71=z-53

Since the first line passes through the point (5, 7, -3) and has direction ratios proportional to 4, -5, -5, its vector equation is
r=a1+λb1                                                      ...(1)               Here,a1=5i^+7j^-3k^ b1=4i^-5j^-5k^

Also, the second line passes through the point (8, 7, 5) and has direction ratios proportional to 7, 1, 3.
Its vector equation is
r=a2+μb2                                                      ... (2)             Here,a2=8i^+7j^+5k^ b2=7i^+j^+3k^

Now,
 a2- a1=3i^+8k^and b1×b2=i^j^k^4-5-5713                    =-10i^-47j^+39k^a2- a1.b1×b2=3i^+8k^.-10i^-47j^+39k^                                =-30+312                                =282We observea2- a1.b1×b20Thus, the given lines do not intersect.

Page No 28.38:

Question 4:

Find the shortest distance between the following pairs of parallel lines whose equations are:
(i) r=i^+2j^+3k^+λi^-j^+k^ and r=2i^-j^-k^+μ-i^+j^-k^

(ii) r=i^+j^+λ2i^-j^+k^ and r=2i^+j^-k^+μ4i^-2j^+2k^

Answer:

(i) The vector equations of the given lines are

r=i^+2j^+3k^+λi^-j^+k^                                                     ...(1)r=2i^-j^-k^+μ-i^+j^-k^   =2i^-j^-k^-μi^-j^+k^                                                 ...(2)

These two lines pass through the points having position vectors a1=i^+2j^+3k^ and a2=2i^-j^-k^ and are parallel to the vector b=i^-j^+k^.

Now,
a2-a1=i^-3j^-4k^
and
a2-a1×b=i^-3j^-4k^×i^-j^+k^                    =i^j^k^1-3-41-11                    =-7i^-5j^+2k^a2-a1×b=-72+-52+22                           =49+25+4                           =78

The shortest distance between the two lines is given by

a2-a1×bb=783=26

(ii) r=i^+j^+λ2i^-j^+k^ and r=2i^+j^-k^+μ4i^-2j^+2k^ or r=2i^+j^-k^+2μ2i^-j^+k^ 

These two lines pass through the points having position vectors a1=i^+j^ and a2=2i^+j^-k^ and are parallel to the vector b=2i^-j^+k^.

Now,
a2-a1=i^-k^
and
a2-a1×b=i^-k^×2i^-j^+k^                    =i^j^k^10-12-11                    =-i^-3j^-k^a2-a1×b=-12+-32+-12                           =1+9+1                           =11

The shortest distance between the two lines is given by

a2-a1×bb=116

Page No 28.38:

Question 5:

Find the equations of the lines joining the following pairs of vertices and then find the shortest distance between the lines
(i) (0, 0, 0) and (1, 0, 2)
(ii) (1, 3, 0) and (0, 3, 0)

Answer:

(i) The equation of the line passing through the points (0, 0, 0) and (1, 0, 2) is

x-01-0=y-00-0=z-02-0=x1=y0=z2

(ii) The equation of the line passing through the points (1, 3, 0) and (0, 3, 0) is

x-10-1=y-33-3=z-00-0=x-1-1=y-30=z0

Since the first line passes through the point (0, 0, 0) and has direction ratios proportional to 1, 0, 2, its vector equation is
r=a1+λb1                                                      ...(1)               Here,a1=0i^+0j^+0k^ b1=i^+0j^+2k^

Also, the second line passes through the point (1, 3, 0) and has direction ratios proportional to -1, 0, 0.
Its vector equation is
r=a2+μb2                                                      ...(2)               Here,a2=i^+3j^+0k^ b2=- i^+0j^+0k^

Now,

 a2- a1=i^+3j^+0k^and b1×b2=i^j^k^102-100=0i^-2j^+0k^ b1×b2=02+-22+02                    =0+4+0                    =2and a2- a1.b1×b2=i^+3j^+0k^.0i^-2j^+0k^                                       =-6

The shortest distance between the lines r=a1+λb1 and r=a2+μb2 is given by

d=a2- a1.b1×b2 b1×b2   =-62   =3  d=3 units

Page No 28.38:

Question 6:

Write the vector equations of the following lines and hence determine the distance between them x-12=y-23=z+46 and x-34=y-36=z+512

Answer:

We have
x-12=y-23=z+46x-34=y-36=z+512

Since the first line passes through the point (1, 2, -4) and has direction ratios proportional to 2, 3, 6, its vector equation is
r=a1+λb1                                                      ...(1)       r=i^+2j^-4k^+λ2i^+3j^+6k^

Also, the second line passes through the point (3, 3, -5) and has direction ratios proportional to 4, 6, 12.
Its vector equation is
r=a2+μb2                                                      ...(2)              r=3i^+3j^-5k^+μ4i^+6j^+12k^r=3i^+3j^-5k^+2μ2i^+3j^+6k^

These two lines pass through the points having position vectors a1=i^+2j^-4k^ and a2=3i^+3j^-5k^ and are parallel to the vector b=2i^+3j^+6k^.

Now,

a2-a1=2i^+j^-k^

and
a2-a1×b=2i^+j^-k^×2i^+3j^+6k^                     =i^j^k^21-1236                     =9i^-14j^+4k^a2-a1×b=92+-142+42                           =81+196+16                           =293and b=22+32+62            =4+9+36            =7

The shortest distance between the two lines is given by

a2-a1×bb=2937units

Page No 28.38:

Question 7:

Find the shortest distance between the lines
(i) r=i^+2j^+k^+λi^-j^+k^ and, r=2i^-j^-k^+μ2i^+j^+2k^

(ii) x+17=y+1-6=z+11 and x-31=y-5-2=z-71

(iii) r=i^+2j^+3k^+λi^-3j^+2k^ and r=4i^+5j^+6k^+μ2i^+3j^+k^

(iv) r=6i^+2j^+2k^+λi^-2j^+2k^ and r=-4i^-k^+μ3i^-2j^-2k^

Answer:

(i) r=i^+2j^+k^+λi^-j^+k^ and r=2i^-j^-k^+μ2i^+j^+2k^

Comparing the given equations with the equations r=a1+λb1 and r=a2+μb2, we get

a1=i^+2j^+k^a2=2i^-j^-k^b1=i^-j^+k^ b2=2i^+j^+2k^ a2- a1=i^-3j^-2k^and b1×b2=i^j^k^1-11212                    =-3i^+3k^ b1×b2=-32+32                    =9+9                    =32and a2- a1.b1×b2=i^-3j^-2k^.-3i^+3k^                                        =-3-6                                        =-9

The shortest distance between the lines r=a1+λb1 and r=a2+μb2 is given by

d=a2- a1.b1×b2 b1×b2d=-932      =32


(ii) x+17=y+1-6=z+11 and x-31=y-5-2=z-71

Since the first line passes through the point (-1, -1, -1) and has direction ratios proportional to 7, -6, 1, its vector equation is
r=a1+λb1                                                                     Here,a1=-i^-j^-k^ b1=7i^-6j^+k^

Also, the second line passing through the point (3, 5, 7) has direction ratios proportional to 1,-2, 1.
Its vector equation is
r=a2+μb2                                                            Here,a2=3i^+5j^+7k^ b2=i^-2j^+k^

Now,
 a2- a1=4i^+6j^+8k^and b1×b2=i^j^k^7-611-21                   =-4i^-6j^-8k^ b1×b2=-42+-62+-82                  =16+36+64                  =116and a2- a1.b1×b2=4i^+6j^+8k^.-4i^-6j^-8k^                                       =-16-36-64                                       =-116

The shortest distance between the lines r=a1+λb1 and r=a2+μb2 is given by

d=a2- a1.b1×b2 b1×b2d=-116116       =116       =229

(iii) r=i^+2j^+3k^+λi^-3j^+2k^ and r=4i^+5j^+6k^+μ2i^+3j^+k^

Comparing the given equations with the equations r=a1+λb1 and r=a2+μb2, we get
a1=i^+2j^+3k^a2=4i^+5j^+6k^b1=i^-3j^+2k^ b2=2i^+3j^+k^ a2- a1=3i^+3j^+3k^and b1×b2=i^j^k^1-32231                    =-9i^+3j^+9k^ b1×b2=-92+32+92                   =81+9+81                   =171and a2- a1.b1×b2=3i^+3j^+3k^.-9i^+3j^+9k^                                       =-27+9+27                                       =9

The shortest distance between the lines r=a1+λb1 and r=a2+μb2 is given by
d=a2- a1.b1×b2 b1×b2d=9171      =319

(iv) r=6i^+2j^+2k^+λi^-2j^+2k^ and r=-4i^-k^+μ3i^-2j^-2k^

Comparing the given equations with the equations r=a1+λb1 and r=a2+μb2, we get
a1=6i^+2j^+2k^a2=-4i^-k^b1=i^-2j^+2k^ b2=3i^-2j^-2k^ a2- a1=-10i^-2j^-3k^and b1×b2=i^j^k^1-223-2-2                    =8i^+8j^+4k^ b1×b2=82+82+42                    =64+64+16                   =144                   =12and a2- a1.b1×b2=-10i^-2j^-3k^.8i^+8j^+4k^                                       =-80-16-12                                       =-108

The shortest distance between the lines r=a1+λb1 and r=a2+μb2 is given by

d=a2- a1.b1×b2 b1×b2d=-10812       =9

Page No 28.38:

Question 8:

Find the distance between the lines l1 and l2 given by
r=i^+2j^-4k^+λ2i^+3j^+6k^ and, r=3i^+3j^-5k^+μ2i^+3j^+6k^

Answer:

Given:
r=i^+2j^-4k^+λ2i^+3j^+6k^ r=3i^+3j^-5k^+μ2i^+3j^+6k^

These two lines pass through the points having position vectors a1=i^+2j^-4k^ and a2=3i^+3j^-5k^ and are parallel to the vector b=2i^+3j^+6k^.

Now,
a2-a1=2i^+j^-k^
and
a2-a1×b=2i^+j^-k^×2i^+3j^+6k^                     =i^j^k^21-1236                     =9i^-14j^+4k^a2-a1×b=92+-142+42                            =81+196+16                           =293and b=22+32+62             =4+9+36             =7

The shortest distance between the two lines is given by

a2-a1×bb=2937



Page No 28.41:

Question 1:

Write the cartesian and vector equations of X-axis.

Answer:

Since x-axis passes through the the point (0, 0, 0) having position vector a=0i^+0j^+0k^ and is parallel to the vector b=1i^+0j^+0k^ having direction ratios proportional to 1, 0, 0, the cartesian equation of x-axis is

x-01=y-00=z-00=x1=y0=z0


Also, its vector equation is

r=a+λb   =0i^+0j^+0k^+λi^+0j^+0k^   =λi^  

Page No 28.41:

Question 2:

Write the cartesian and vector equations of Y-axis.

Answer:

Since y-axis passes through the the point (0, 0, 0) having position vector a=0i^+0j^+0k^ and is parallel to the vector b=0i^+1j^+0k^ having direction ratios proportional to 0, 1, 0, the cartesian equation of y-axis is

x-00=y-01=z-00=x0=y1=z0


Also, its vector equation is

r=a+λb   =0i^+0j^+0k^+λ0i^+j^+0k^   =λj^  

Page No 28.41:

Question 3:

Write the cartesian and vector equations of Z-axis.

Answer:

Since z-axis passes through the the point (0, 0, 0) having position vector a=0i^+0j^+0k^ and is parallel to the vector b=0i^+0j^+k^ having direction ratios proportional to 0, 0, 1, the cartesian equation of z-axis is

x-00=y-00=z-01=x0=y0=z1

Also, its vector equation is

r=a+λb   =0i^+0j^+0k^+λ0i^+0j^+k^   =λk^  

Page No 28.41:

Question 4:

Write the vector equation of a line passing through a point having position vector α and parallel to vector β .

Answer:

The vector equation of the line passing through the point having position vector α and parallel to vector β is r=α+λβ.

Page No 28.41:

Question 5:

Cartesian equations of a line AB are 2x-12=4-y7=z+12. Write the direction ratios of a line parallel to AB.

Answer:

We have

2x-12=4-y7=z+12

The equation of the line AB can be re-written as

x-121=y-4-7=z+12

The direction ratios of the line parallel to AB are proportional to 1, -7, 2.

Also, the direction cosines of the line parallel to AB are proportional to

112+-72+22, -712+-72+22, 212+-72+22 =154, -754, 254

Page No 28.41:

Question 6:

Write the direction cosines of the line whose cartesian equations are 6x − 2 = 3y + 1 = 2z − 4.

Answer:

We have
6x − 2 = 3y + 1 = 2z − 4

The equation of given line can be re-written as

x-1316=y+1313=z-212x-131=y+132=z-23

The direction ratios of the line parallel to AB are proportional to 1, 2, 3.

Hence, the direction cosines of the line parallel to AB are proportional to

112+22+32, 212+22+32, 312+22+32=114, 214, 314.

Page No 28.41:

Question 7:

Write the direction cosines of the line x-22=2y-5-3, z=2.

Answer:

We have

x-22=2y-5-3, z=2

The equation of the given line can be re-written as

x-22=y-52-32= z-20x-24=y-52-3= z-20


The direction ratios of the given line are proportional to 4, -3, 0.

Hence, the direction cosines of the given line are proportional to

442+-32+02, -342+-32+02, 042+-32+02=45, -35, 0

Page No 28.41:

Question 8:

Write the coordinate axis to which the line x-23=y+14=z-10is perpendicular.

Answer:

We have

x-23=y+14=z-10

The given line is parallel to the vector b=3i^+4j^+0k^.

Let xi^+yj^+zk^ be perpendicular to the given line.

Now,
3x+4y+0z=0
It is satisfied by the coordinates of z-axis, i.e. (0, 0, 1).

Hence, the given line is perpendicular to z-axis.

Page No 28.41:

Question 9:

Write the angle between the lines x-57=y+2-5=z-21 and x-11=y2=z-13.

Answer:

We have

x-57=y+2-5=z-21 x-11=y2=z-13

The given lines are parallel to the vectors b1=7i^-5j^+k^ and b2=i^+2j^+3k^.

Let θ be the angle between the given lines.

Now,
cos θ=b1.b2b1 b2         =7i^-5j^+k^.i^+2j^+3k^72+-52+12 12+22+32         =7-10+349+25+1 1+4+9         =0θ=π2

Page No 28.41:

Question 10:

Write the direction cosines of the line whose cartesian equations are 2x = 3y = −z.

Answer:

We have
2x = 3y = −z

The equation of the given line can be re-written as

x12=y13=z-1x3=y2=z-6

The direction ratios of the line parallel to AB are proportional to 3, 2, -6.

Hence, the direction cosines of the line parallel to AB are proportional to

332+22+-62, 232+22+-62, -632+22+-62=37, 27, -67

Page No 28.41:

Question 11:

Write the angle between the lines 2x = 3y = −z and 6x = −y = −4z.

Answer:

We have
2x = 3y = −z
6x = −y = −4z

The given lines can be re-written as

x12=y13=z-1 and x16=y-1=z-14x3=y2=z-6 and x2=y-12=z-3

These lines are parallel to vectors b1=3i^+2j^-6k^ and b2=2i^-12j^-3k^.

Let θ be the angle between these lines.

Now,

cos θ=b1.b2b1 b2=3i^+2j^-6k^.2i^-12j^-3k^32+22+-62 22+-122+-32=6-24+189+4+36 4+144+9=0θ=π2

Page No 28.41:

Question 12:

Write the value of λ for which the lines x-3-3=y+22λ=z+42 and x+13λ=y-21=z+6-5 are perpendicular to each other.

Answer:

We have

x-3-3=y+22λ=z+42 x+13λ=y-21=z+6-5

The given lines are parallel to vectors b1=-3i^+2λj^+2k^ and b2=3λi^+j^-5k^.

For b1b2, we must have

b1.b2=0-3i^+2λj^+2k^.3λi^+j^-5k^=0-7λ-10=0λ=-107

Page No 28.41:

Question 13:

Write the formula for the shortest distance between the lines r=a1+λb and r=a2+μb.

Answer:

The shortest distance d between the parallel lines r=a1+λb and r=a2+μb is given by

d=a2-a1×bb

Page No 28.41:

Question 14:

Write the condition for the lines r=a1+λb1 and r=a2+μb2 to be intersecting.

Answer:

The shortest distance between the lines r=a1+λb1 and r=a2+λb2 is given by

d=a2- a1.b1×b2 b1×b2

For the lines to be intersecting, d=0.

a2- a1.b1×b2 b1×b2=0a2- a1.b1×b2=0

Page No 28.41:

Question 15:

The cartesian equations of a line AB are 2x-13=y+22=z-33. Find the direction cosines of a line parallel to AB.

Answer:

We have

2x-13=y+22=z-33

The equation of the line AB can be re-written as

x-1232=y+22=z-33=x-123=y+24=z-36

Thus, the direction ratios of the line parallel to AB are proportional to 3, 4, 6.

Hence, the direction cosines of the line parallel to AB are proportional to

332+42+62, 432+42+62, 632+42+62 = 355, 455, 655

Page No 28.41:

Question 16:

If the equations of a line AB are 3-x1=y+2-2=z-54, write the direction ratios of a line parallel to AB.

Answer:

We have

3-x1=y+2-2=z-54

The equation of the line AB can be re-written as

x-3-1=y+2-2=z-54

Thus, the direction ratios of the line parallel to AB are proportional to -1, -2, 4.

Page No 28.41:

Question 17:

Write the vector equation of a line given by x-53=y+47=z-62.

Answer:

We have

x-53=y+47=z-62

The given line passes through the point (5, -4, 6) and has direction ratios proportional to 3, 7, 2.

Vector equation of the given line passing through the point having position vector a=5i^-4j^+6k^ and parallel to a vector b=3i^+7j^+2k^ is

r=a+λbr=5i^-4j^+6k^+λ3i^+7j^+2k^

Disclaimer: The answer given in the book is incorrect. This solution is created according to the question given in the book.



Page No 28.42:

Question 18:

The equations of a line are given by 4-x3=y+33=z+26. Write the direction cosines of a line parallel to this line.

Answer:

We have

4-x3=y+33=z+26

The equation of the given line can be re-written as

x-4-3=y+33=z+26

The direction ratios of the line parallel to the given line are proportional to -3, 3, 6.


Hence, the direction cosines of the line parallel to the given line are proportional to

-3-32+32+62, 3-32+32+62, 6-32+32+62 =-16, 16, 26

Page No 28.42:

Question 19:

Find the Cartesian equations of the line which passes through the point (−2, 4 , −5) and is parallel to the line x+33=4-y5=z+86.

Answer:

The equation of the given line is x+33=4-y5=z+86

It can be re-written as

x+33=y-4-5=z+86

Since the required line is parallel to the given line, the direction ratios of the required line are proportional to 3, -5, 6.

Hence, the cartesian equations of the line passing through the point (-2, 4, -5) and parallel to a vector having direction ratios proportional to 3, -5, 6 is x+23=y-4-5=z+56.

Page No 28.42:

Question 20:

Find the angle between the lines r=2i^-5j^+k^+λ3i^+2j^+6k^ and r=7i^-6k^+μi^+2j^+2k^.                  [CBSE 2014]

Answer:


Let θ be the angle between the given lines. The given lines are parallel to the vectors b1=3i^+2j^+6k^ and b2=i^+2j^+2k^, respectively.

So, the angle θ between the given lines is given by

cosθ=b1.b2b1b2        =3i^+2j^+6k^.i^+2j^+2k^32+22+6212+22+22        =3×1+2×2+6×2499        =1921θ=cos-11921

Thus, the angle between the given lines is cos-11921.

Page No 28.42:

Question 21:

Find the angle between the lines 2x=3y=-z and 6x=-y=-4z.           [CBSE 2015]

Answer:


The equations of the given lines can be re-written as

x3=y2=z-6 and x2=y-12=z-3

We know that angle between the lines x-x1a1=y-y1b1=z-z1c1 and x-x2a2=y-y2b2=z-z2c2 is given by cosθ=a1a2+b1b2+c1c2a12+b12+c12a22+b22+c22.

Let θ be the angle between the given lines.

cosθ=3×2+2×-12+-6×-332+22+-6222+-122+-32            =6-24+1849157            =0θ=π2

Thus, the angle between the given lines is π2.

Page No 28.42:

Question 1:

The angle between the straight lines
x+12=y-25=z+34 and x-11=y+22=z-3-3 is
(a) 45°
(b) 30°
(c) 60°
(d) 90°

Answer:

(d) 90°

We have

x+12=y-25=z+34 x-11=y+22=z-3-3

The direction ratios of the given lines are proportional to 2, 5, 4 and 1, 2, -3.

The given lines are parallel to the vectors b1=2i^+5j^+4k^ and b2=i^+2j^-3k^.

Let θ be the angle between the given lines.

Now,
cos θ=b1.b2b1 b2         =2i^+5j^+4k^.i^+2j^-3k^22+52+42 12+22+-32         =2+10-1245 14         =0θ=90°

Page No 28.42:

Question 2:

The lines x1=y2=z3 and x-1-2=y-2-4=z-3-6 are
(a) coincident
(b) skew
(c) intersecting
(d) parallel

Answer:

(a) coincident

The equation of the given lines are

x1=y2=z3                          ...(1)                


x-1-2=y-2-4=z-3-6=x-11=y-22=z-33             ...(2)

Thus, the two lines are parallel to the vector b=i^+2j^+3k^ and pass through the points (0, 0, 0) and (1, 2, 3).

Now,

a2-a1×b=i^+2j^+3k^×i^+2j^+3k^                   =0          a×a=0

Since the distance between the two parallel line is 0, the given lines are coincident.

Page No 28.42:

Question 3:

The direction ratios of the line perpendicular to the lines

x-72=y+17-3=z-61 and, x+51=y+32=z-4-2 are proportional to
(a) 4, 5, 7
(b) 4, −5, 7
(c) 4, −5, −7
(d) −4, 5, 7

Answer:

(a) 4, 5, 7

We have

x-72=y+17-3=z-61 x+51=y+32=z-4-2

The direction ratios of the given lines are proportional to 2, -3, 1 and 1, 2, -2.

The vectors parallel to the given vectors are b1=2i^-3j^+k^ and b2=i^+2j^-2k^.

Vector perpendicular to the given two lines is
b=b1×b2  =i^j^k^2-3112-2  =4i^+5j^+7k^

Hence, the direction ratios of the line perpendicular to the given two lines are proportional to 4, 5, 7.

Page No 28.42:

Question 4:

The angle between the lines x-11=y-11=z-12 and, x-1-3-1=y-13-1=z-14 is
(a) cos-1165

(b) π6

(c) π3

(d) π4

Answer:

(c) π3

We have

x-11=y-11=z-12 x-1-3-1=y-13-1=z-14


The direction ratios of the given lines are proportional to 1, 1, 2 and -3-1, 3-1, 4.

The given lines are parallel to vectors b1=i^+j^+2k^ and b2=-3-1i^+3-1j^+4k^.

Let θ be the angle between the given lines.

Now,

cos θ=b1.b2b1 b2=i^+j^+2k^.-3-1i^+3-1j^+4k^12+12+12 -3-12+3-12+42=-3-1+3-1+83 24=662=12θ=π3



Page No 28.43:

Question 5:

The direction ratios of the line xy + z − 5 = 0 = x − 3y − 6 are proportional to
(a) 3, 1, −2

(b) 2, −4, 1

(c) 314, 114, -214

(d) 241, -441, 141

Answer:

(a) 3, 1, −2

We have
xy + z − 5 = 0 = x − 3y − 6

x-3y-6=0      x-y+z-5=0x=3y+6                 ... 1      x-y+z-5=0       ... 2

From (1) and (2), we get

3y+6-y+z-5=02y+z+1=0y=-z-12 y=x-63    From 1 x-63=y=-z-12

So, the given equation can be re-written as

x-63=y1=z+1-2

Hence, the direction ratios of the given line are proportional to 3, 1,-2.

Page No 28.43:

Question 6:

The perpendicular distance of the point P (1, 2, 3) from the line x-63=y-72=z-7-2 is
(a) 7
(b) 5
(c) 0
(d) none of these

Answer:

(a) 7

We have

x-63=y-72=z-7-2

Let point (1, 2, 3) be P and the point through which the line passes be Q (6, 7, 7). Also, the line is parallel to the vector b=3i^+2j^-2k^.

Now,

PQ=5i^+5j^+4k^

 b×PQ=i^j^k^32-2554                 =18i^-22j^+5k^b×PQ=182+-222+52                   =324+484+25                   =833 d=b×PQb         =83317         =49         =7

Page No 28.43:

Question 7:

The equation of the line passing through the points a1 i^+a2 j^+a3 k^ and b1 i^+b2 j^+b3 k^ is
(a) r=a1 i^+a2 j^+a3 k^+λ b1 i^+b2 j^+b3 k^

(b) r=a1 i^+a2 j^+a3 k^-t b1 i^+b2 j^+b3 k^

(c) r=a1 1-t i^+a2 1-t j^+a3 1-t k^+t b1 i^+b2 j^+b3 k^

(d) none of these

Answer:

(c) r=a1 1-t i^+a2 1-t j^+a3 1-t k^+t b1 i^+b2 j^+b3 k^

Equation of the line passing through the points having position vectors a1 i^+a2 j^+a3 k^ and b1 i^+b2 j^+b3 k^ is

r=a1 i^+a2 j^+a3 k^+tb1 i^+b2 j^+b3 k^-a1 i^+a2 j^+a3 k^, where t is a parameter   =a1 i^+a2 j^+a3 k^-ta1 i^+a2 j^+a3 k^+tb1 i^+b2 j^+b3 k^   =a11-t i^+a21-t j^+a31-t k^+tb1 i^+b2 j^+b3 k^

Page No 28.43:

Question 8:

If a line makes angles α, β and γ with the axes respectively, then cos 2 α + cos 2 β + cos 2 γ =
(a) −2
(b) −1
(c) 1
(d) 2

Answer:

(b) −1
 
If a line makes angles α, β and γ with the axes, then
cos2α+cos2β+cos2γ=1                           ...(1)

We have
cos 2α+cos 2β+ cos 2γ=2 cos2α-1+2 cos2β-1+2 cos2γ-1         cos 2θ=2 cos2θ-1                                             =2 cos2α+cos2β+cos2γ-3                 From 1                                             =21-3                                             =-1

Page No 28.43:

Question 9:

If the direction ratios of a line are proportional to 1, −3, 2, then its direction cosines are
(a) 114, -314, 214

(b) 114, 214, 314

(c) -114, 314, 214

(d) -114, -214, -314

Answer:

(a) 114, -314, 214
 
 
The direction ratios of the line are proportional to 1, -3, 2.

The direction cosines of the line are

112+-32+22, -312+-32+22, 212+-32+22 =114, -314, 214

Page No 28.43:

Question 10:

If a line makes angle π3 and π4 with x-axis and y-axis respectively, then the angle made by the line with z-axis is
(a) π/2
(b) π/3
(c) π/4
(d) 5π/12

Answer:

(b) π/3

If a line makes angles α, β and γ with the axes, then cos2α+cos2β+cos2γ=1.

Here,

α=π3β=π4

Now,

cos2α+cos2β+cos2γ=1cos2π3+cos2π4+cos2γ=114+12+cos2γ=1cos2γ=1-34cos2γ=14cos γ=12γ=π3

Page No 28.43:

Question 11:

The projections of a line segment on X, Y and Z axes are 12, 4 and 3 respectively. The length and direction cosines of the line segment are
(a) 13; 1213, 413, 313

(b) 19; 1219, 419, 319

(c) 11; 1211, 1411, 311

(d) none of these

Answer:

(a) 13; 1213, 413, 313

If a line makes angles α, β and γ with the axes, then
cos2α+cos2β+cos2γ=1                               ...(1)                  

Let r be the length of the line segment. Then,

r cos α=12, r cos β=4, r cos γ=3                    ...(2)r cos α2+r cos β2+r cos γ2=122+42+32r2 cos2 α+cos2 β+cos2 γ=169r2 1=169           From 1r=169r=±13r=13                      (Since length cannot be negative)


Substituting r = 13 in (2), we get

cos α=1213, cos β=413, cos γ=113

Thus, the direction cosines of the line are 1213, 413, 113.

Page No 28.43:

Question 12:

The lines x1=y2=z3 and x-1-2=y-2-4=z-3-6 are
(a) parallel
(b) intersecting
(c) skew
(d) coincident

Answer:

(d) coincident

The equations of the given lines are

x1=y2=z3                             ...(1)                      


x-1-2=y-2-4=z-3-6x-11=y-22=z-33           ...2

Thus, the two lines are parallel to the vector b=i^+2j^+3k^ and pass through the points (0, 0, 0) and (1, 2, 3).

Now,

a2-a1×b=i^+2j^+3k^×i^+2j^+3k^                   =0         a×a=0

Since, the distance between the two parallel lines is 0, the given two lines are coincident lines.


Disclaimer: The answer given in the book is incorrect. This solution is created according to the question given in the book.

Page No 28.43:

Question 13:

The straight line x-33=y-21=z-10 is
(a) parallel to x-axis
(b) parallel to y-axis
(c) parallel to z-axis
(d) perpendicular to z-axis

Answer:

(d) perpendicular to z-axis

We have

x-33=y-21=z-10

Also, the given line is parallel to the vector b=3i^+j^+0k^.

Let xi^+yj^+zk^ be perpendicular to the given line.

Now,
3x+4y+0z=0
It is satisfied by the coordinates of z-axis, i.e. (0, 0, 1).

Hence, the given line is perpendicular to z-axis.

Page No 28.43:

Question 14:

The shortest distance between the lines
x-33=y-8-1=z-31 and, x+3-3=y+72=z-64 is
(a) 30
(b) 230
(c) 530
(d) 330

Answer:

(d) 330

We have

x-33=y-8-1=z-31               ...(1) x+3-3=y+72=z-64              ...(2)

We know that line (1) passes through the point (3, 8, 3) and has direction ratios proportional to 3, -1, 1.

Its vector equation is r=a1+λb1, where a1=3i^+8j^+3k^ and b1=3i^-j^+k^.

Also, line (2) passes through the point (-3, -7, 6) and has direction ratios proportional to -3, 2, 4.

Its vector equation is r=a2+μb2, where a2=-3i^-7j^+6k^ and b2=-3i^+2j^+4k^.

Now,

 a2- a1=-6i^-15j^+3k^b1×b2=i^j^k^3-11-324            =-6i^-15j^+3k^ b1×b2=-62+-152+32                   =36+225+9                   =270a2- a1.b1×b2=-6i^-15j^+3k^.-6i^-15j^+3k^                                =36+225+9                                =270

The shortest distance between the lines r=a1+λb1 and r=a2+μb2 is given by

d=a2- a1.b1×b2 b1×b2  =270270  =270  =330



Page No 28.9:

Question 1:

Find the vector and cartesian equations of the line through the point (5, 2, −4) and which is parallel to the vector 3i^+2j^-8k^.

Answer:

We know that the vector equation of a line passing through a point with position vector a and parallel to vector b is r = a + λ b.

Here,
a = 5i^+2j^-4k^b =3i^+2j^-8k^ 

Vector equation of the required line is given by
r = 5i^+2j^-4k^ + λ 3i^+2j^-8k^                                   ...1 Here, λ is a parameter.

Reducing (1) to cartesian form, we get

xi^+yj^+zk^ = 5i^+2j^-4k^ + λ 3i^+2j^-8k^                    [Putting r = xi^+yj^+zk^  in (1)]xi^+yj^+zk^ = 5+3λ i^+2+2 λ j^+-4-8 λ k^Comparing the coefficients of i^,  j^ and k^, we getx=5+3λ, y=2+2 λ, z=-4-8 λx-53=λ, y-22=λ, z+4-8=λx-53=y-22=z+4-8=λHence, the cartesian form of (1) isx-53=y-22=z+4-8

Page No 28.9:

Question 2:

Find the vector equation of the line passing through the points (−1, 0, 2) and (3, 4, 6).

Answer:

We know that the vector equation of a line passing through the points with position vectors a and b is r = a + λ b-a, where λ is a scalar.

Here,
a =-1i^+0j^+2k^b= 3i^+4j^+6k^ 

Vector equation of the required line is
r=-1i^+0j^+2k^ + λ 3i^+4j^+6k^--1i^+0j^+2k^r=-1i^+0j^+2k^ + λ 4i^+4j^+4k^Here, λ is a parameter.

Page No 28.9:

Question 3:

Find the vector equation of a line which is parallel to the vector 2i^-j^+3k^ and which passes through the point (5, −2, 4). Also, reduce it to cartesian form.

Answer:

We know that the vector equation of a line passing through a point with position vector a and parallel to the vector b is r = a + λ b.

Here,
a = 5i^-2j^+4k^b = 2i^-j^+3k^ 

So, the vector equation of the required line is
r = 5i^-2j^+4k^ + λ 2i^-j^+3k^                                   ...(1) Here, λ is a parameter.

Reducing (1) to cartesian form, we get

xi^+yj^+zk^= 5i^-2j^+4k^ +λ 2i^-j^+3k^                     [Putting r=xi^+yj^+zk^ in (1)]xi^+yj^+zk^ = 5+2λ i^+-2-λ j^+4+3λ k^Comparing the coefficients of  i^,  j^ and k^, we getx=5+2λ, y=-2-λ, z=4+3λx-52=λ, y+2-1=λ, z-43=λx-52=y+2-1=z-43=λHence, the cartesian form of (1) isx-52=y+2-1=z-43

Page No 28.9:

Question 4:

A line passes through the point with position vector 2i^-3j^+4k^ and is in the direction of 3i^+4j^-5k^. Find equations of the line in vector and cartesian form.

Answer:

We know that the vector equation of a line passing through a point with position vector a and parallel to the vector b is r = a + λ b.

Here,
a = 2i^-3j^+4k^ b = 3i^+4j^-5k^.

So, the vector equation of the required line is
r = 2i^-3j^+4k^ + λ 3i^+4j^-5k^                                   ...(1)Here, λ is a parameter.

Reducing (1) to cartesian form, we get

xi^+yj^+zk^ =  2i^-3j^+4k^ + λ 3i^+4j^-5k^                  [Putting r = xi^+yj^+zk^ in (1)]xi^+yj^+zk^ = 2+3λ i^+-3+4λ j^+4-5λ k^Comparing the coefficients of  i^,  j^ and k^, we getx=2+3λ, y=-3+4λ, z=4-5λx-23=λ, y+34=λ, z-4-5=λx-23=y+34=z-4-5=λHence, the cartesian form of (1) isx-23=y+34=z-4-5

Page No 28.9:

Question 5:

ABCD is a parallelogram. The position vectors of the points A, B and C are respectively, 4i^+5j^-10k^, 2i^-3j^+4k^ and -i^+2j^+k^. Find the vector equation of the line BD. Also, reduce it to cartesian form.

Answer:

We know that the position vector of the mid-point of a and b is a+b2.

Let the position vector of point D be xi^+yj^+zk^.

Position vector of mid-point of A and C = Position vector of mid-point of B and D

 4i^+5j^-10k^+-i^+2j^+k^2=2i^-3j^+4k^+xi^+yj^+zk^232i^+72j^-92k^=x+22i^+-3+y2j^+4+z2k^Comparing the coefficient of i^,  j^ and k^, we getx+22=32x=1-3+y2=72y=10 4+z2=-92 z=-13Position vector of point D=i^+10j^-13k^

The vector equation of line BD passing through the points with position vectors a(B) and b(D) is r = a + λ b-a.

Here,
a = 2i^-3j^+4k^ b =i^+10j^-13k^

Vector equation of the required line is
r = 2i^-3j^+4k^+λi^+10j^-13k^-2i^-3j^+4k^r = 2i^-3j^+4k^ + λ -i^+13j^-17k^                                   ...(1) Here, λ is a parameter.

Reducing (1) to cartesian form, we get

xi^+yj^+zk^ = 2i^-3j^+4k^ + λ -i^+13j^-17k^                     [Puttingr = xi^+yj^+zk^ in (1)]xi^+yj^+zk^ = 2-λ i^+-3+13λ j^+4-17λ k^Comparing the coefficients of  i^,  j^ and k^, we getx=2-λ, y=-3+13λ, z=4-17λx-2-1=λ, y+313=λ, z-4-17=λx-2-1=y+313=z-4-17=λx-21=y+3-13=z-417=-λHence, the cartesian form of (1) isx-21=y+3-13=z-417

Page No 28.9:

Question 6:

Find in vector form as well as in cartesian form, the equation of the line passing through the points A (1, 2, −1) and B (2, 1, 1).

Answer:

We know that the vector equation of a line passing through the points with position vectors a and b is r = a + λ b-a, where λ is a scalar.

Here,
a = i^+2j^-k^ b =2i^+j^+k^

Vector equation of the required line is
r = i^+2j^-k^+λ2i^+j^+k^-i^+2j^-k^r = i^+2j^-k^ + λi^-j^+2k^                                   ...(1) Here, λ is a parameter.

Reducing (1) to cartesian form, we get

xi^+yj^+zk^ = i^+2j^-k^ + λi^-j^+2k^               [Putting r=xi^+yj^+zk^ in (1)]xi^+yj^+zk^ = 1+λ i^+2-λ j^+-1+2λ k^Comparing the coefficients of  i^,  j^ and k^, we getx=1+λ, y=2-λ, z=-1+2λx-1=λ, y-2-1=λ, z+12=λx-11=y-2-1=z+12=λHence, the cartesian form of (1) isx-11=y-2-1=z+12

Page No 28.9:

Question 7:

Find the vector equation for the line which passes through the point (1, 2, 3) and parallel to the vector i^-2j^+3k^. Reduce the corresponding equation in cartesian from.

Answer:

We know that the vector equation of a line passing through a point with position vector a and parallel to the vector b is r = a + λ b.

Here,
a =i^+2j^+3k^ b =i^-2j^+3k^

Vector equation of the required line is
r = i^+2j^+3k^ + λ i^-2j^+3k^                                   ... (1)Here, λ is a parameter.

Reducing (1) to cartesian form, we get

xi^+yj^+zk^ =  i^+2j^+3k^ + λ i^-2j^+3k^                       [Putting r=xi^+yj^+zk^  in (1)]xi^+yj^+zk^ = 1+λ i^+2-2λ j^+3+3λ k^Comparing the coefficients of  i^,  j^ and k^, we getx=1+λ, y=2-2λ, z=3+3λx-1=λ, y-2-2=λ, z-33=λx-11=y-2-2=z-33=λHence, the cartesian form of (1)isx-11=y-2-2=z-33



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