RD Sharma XII Vol 2 2018 Solutions for Class 12 Science Math Chapter 8 Direction Cosines And Direction Ratios are provided here with simple step-by-step explanations. These solutions for Direction Cosines And Direction Ratios are extremely popular among class 12 Science students for Math Direction Cosines And Direction Ratios Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma XII Vol 2 2018 Book of class 12 Science Math Chapter 8 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RD Sharma XII Vol 2 2018 Solutions. All RD Sharma XII Vol 2 2018 Solutions for class 12 Science Math are prepared by experts and are 100% accurate.

Page No 27.23:

Question 1:

If a line makes angles of 90°, 60° and 30° with the positive direction of x, y, and z-axis respectively, find its direction cosines.

Answer:

Let the direction cosines of the line be l, m, n.

Now,

l = cos900 = 0m = cos600 = 12n = cos300 = 32Therefore, the direction cosines of the line are 0, 12, 32.

Page No 27.23:

Question 2:

If a line has direction ratios 2, −1, −2, determine its direction cosines.

Answer:

Let the direction cosines of the line be l, m, n.Now, l = 222+-12+-22=23m = -122+-12+-22=-13n =-222+-12+-22=-23Therefore, the direction cosines of the line are 23 ,-13, -23.

Page No 27.23:

Question 3:

Find the direction cosines of the line passing through two points (−2, 4, −5) and (1, 2, 3).

Answer:

The direction cosines of the line passing through two points P x1, y1, z1 and Q x2, y2, z2 are x2-x1PQ, y2-y1PQ,z2-z1PQ.Here,PQ=x2-x12 + y2-y12 + z2-z12P=-2, 4,-5 Q=1,2,3 PQ = 1--22 + 2-42 + 3 - -52 =77Thus, the direction cosines of the line joining two points are 1--277, 2-477, 3 - -577, i.e. 377, -277, 877.

Page No 27.23:

Question 4:

Using direction ratios show that the points A (2, 3, −4), B (1, −2, 3) and C (3, 8, −11) are collinear.

Answer:

The given points are A 2, 3, -4, B1, -2, 3 and C 3, 8, -11.We know that the direction ratios of the line joining the points, x1, y1, z1 and x2, y2, z2 are x2-x1, y2-y1, z2-z1.The direction ratios of the line joining A and B are 1-2, -2-3, 3+4, i.e.-1, -5,  7.The direction ratios of the line joining B and C are 3-1, 8+2, -11-3, i.e. 2, 10, -14.It is clear that the direction ratios of BC are -2 times that of AB, i.e. they are proportional.Therefore, AB is parallel to BC. Also, point B is common in both AB and BC. Therefore, points A, B and C are collinear.

Page No 27.23:

Question 5:

Find the direction cosines of the sides of the triangle whose vertices are (3, 5, −4), (−1, 1, 2) and (−5, −5, −2).

Answer:

The vertices of ABC are A 3, 5, -4, B -1,1,2 and C -5, -5, -2.



The direction ratios of AB are -1-3, 1-5, 2--4, i.e. -4, -4, 6.Therefore, the direction cosines of AB are-4-42+-42+62, -4-42+-42+62, 6-42+-42+62=-4 217, -4217, 6217  = 217, 217, -317The direction ratios of BC are -5--1, -5-1, -2-2, i.e. -4, -6,-4.Therefore, the direction cosines of BC are-4-42+-62+-42, -6-42+-62+-42, -4-42+-62+-42=-4217, -6217, -4217 =217, 37, 217The direction ratios of CA are 3--5, 5--5, -4--2, i.e. 8, 10, -2.Therefore, the direction cosines of CA are882+102+-22, 1082+102+-22, -282+102+-22=8242, 10242, -2242 =442, 542, -142

Page No 27.23:

Question 6:

Find the angle between the vectors with direction ratios proportional to 1, −2, 1 and 4, 3, 2.

Answer:

Let a be a vector with direction ratios 1, -2, 1.a  = i^ -2j^ + k^ . Let b be a vector with direction ratios 4, 3, 2.b=4i^ + 3j^ + 2k^.Let θ be the angle between the given vectors. Now,cos θ = a.ba b           =i^ -2j^ + k^.4i^ + 3j^ + 2k^i^ -2j^ + k^4i^ + 3j^ + 2k^          = 4-6+21+4+1 16+9+4           = 06 29          = 0  θ =π2Thus, the angle between the given vectors measures π2.

Page No 27.23:

Question 7:

Find the angle between the vectors whose direction cosines are proportional to 2, 3, −6 and 3, −4, 5.

Answer:

Let a be a vector with direction ratios 2, 3, -6.  a = 2i^ +3j^ -6k^.Let b be a vector with direction ratios 3, -4, 5.b = 3i^ -4j^ + 5k^Let θ be the angle between the given vectors. Now,cos θ = a. ba b           =2i^ +3j^ -6k^.3i^ -4j^ + 5k^2i^ +3j^ -6k^3i^ -4j^ + 5k^          = 6-12-304+9+36 9+16+25            = -3649 50           =-36352Rationalising the result, we getcos θ = -18235  θ = cos-1-18235Thus, the angle between the given vectors measures cos-1-18235.

Page No 27.23:

Question 8:

Find the acute angle between the lines whose direction ratios are proportional to 2 : 3 : 6 and 1 : 2 : 2.

Answer:

Let a be a vector parallel to the vector with direction ratios 2, 3, 6. a=2i^+3j^+6k^. Let b be a vector parallel to the vector with direction ratios 1, 2, 2.b= i^ +2j^ +2k^Let θ be the angle between the the given vectors. Now,cos θ =a.ba b          =2i^+3j^+6k^.i^ +2j^ +2k^2i^+3j^+6k^i^ +2j^ +2k^         = 2+6+12 4+9+36 1+4+4         = 2021 θ=cos-12021

Page No 27.23:

Question 9:

Show that the points (2, 3, 4), (−1, −2, 1), (5, 8, 7) are collinear.

Answer:

Suppose the points are A 2,3,4, B -1.-2,1 and C 5,8,7.We know that the direction ratios of the line joining the points x1, y1, z1 and x2, y2, z2 are x2-x1, y2-y1, z2-z1.The direction ratios of AB are -1-2, -2-3, 1-4, i.e. -3, -5,-3.The direction ratios of BC are 5--1, 8--2, 7-1, i.e. 6, 10, 6.It can be seen that the direction ratios of BC are -2 times that of AB, i.e. they are proportional. Therefore, AB is parallel to BC. Since point B is common in both AB and BC, points A, B, and C are collinear.

Page No 27.23:

Question 10:

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

Answer:

We know that the direction ratios of the line passing through the points x1, y1, z1 and x2, y2, z2 are x2-x1, y2-y1, z2-z1.Let the first two points be A 4, 7, 8 and B 2, 3, 4.Thus, the direction ratios of AB are 2-4, 3-7, 4-8, i.e. -2, -4, -4.Similarly, let the other two points be C -1,-2, 1 and D1, 2, 5.Thus, the direction ratios of CD are 1--1, 2--2, 5-1, i.e. 2, 4, 4.It can be seen that the direction ratios of CD are-1 times that of AB, i.e. they are proportional.Therefore, AB and CD are parallel lines.

Page No 27.23:

Question 11:

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

Answer:

We know that two lines with direction ratios a1, b1, c1 and a2, b2, c2 are perpendicular if a1a2 + b1b2+c1c2=0.The direction ratios of the line passing through the points 1, -1, 2 and 3, 4, -2 are 3-1, 4--1, -2-2, i.e. 2, 5,-4.a1=2,  b1=5, c1 = -4Similarly, the direction ratios of the line passing through the points 0, 3, 2 and 3, 5, 6 are 3-0, 5-3, 6-2, i.e. 3, 2, 4.a2=3, b2=2, c2=4 a1a2 + b1b2+c1c2 = 2×3 + 5×2 +-4×4 =6 +10 - 16 = 0

Thus, the line through the points (1, -1, 2) and (3, 4, -2) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6).

Page No 27.23:

Question 12:

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:

We know that two lines with direction ratios a1, b1, c1 and a2, b2, c2 are perpendicular if a1a2+b1b2+c1c2=0.The direction ratios of the line joining the origin 0, 0, 0 to the point 2, 1, 1 are 2-0, 1-0, 1-0 or 2, 1, 1.a1=2,  b1=1, c1=1Similarly, the direction ratios of the line joining the points 3, 5, -1 and 4, 3, -1 are 4-3, 3-5, -1--1 or 1, -2, 0.a2=1, b2=-2, c2=0 a1a2+b1b2+c1c2= 2 -2 + 0=0

Therefore, 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).

Page No 27.23:

Question 13:

Find the angle between the lines whose direction ratios are proportional to a, b, c and bc, ca, ab.

Answer:

Let θ be the angle between the given lines.We have   a1=a, b1=b, c1=c  a2=b-c, b2=c-a, c2=a-bNow,cos θ = a1a2+b1b2+c1c2a12+b12+c12a22+b22+c22           =ab-c+bc-a+ca-ba2+b2+c2b-c2+c-aa-b=ab-ac+bc-ab+ac-bca2+b2+c2b-c2+c-aa-b=0θ=π2Thus, the angle between the given lines measures 90°.

Page No 27.23:

Question 14:

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

Answer:

The given points are A 1, 2, 3, B4, 5, 7, C-4, 3, -6 and D 2, 9, 2.We know that the direction ratios of the line joining the points x1, y1, z1 and x2, y2, z2 are x2-x1, y2-y1, z2-z1.The direction ratios of AB are 4-1, 5-2, 7-3, i.e. 3, 3, 4.The direction ratios of CD are 2--4, 9-3, 2--6, i.e. 6, 6, 8.Let θ be the angle between AB and CD.We havea1=3, b1=3, c1=4 a2=6, b2=6, c2=8 cos θ =a1a2+b1b2+c1c2a12+b12+c12a22+b22+c22=18+18+329+9+1636+36+64=6868=1 θ =0°Thus, the angle between AB and CD measures 0°.

Page No 27.23:

Question 15:

Find the direction cosines of the lines, connected by the relations: l + m +n = 0 and 2lm + 2lnmn = 0.

Answer:

Given:l+m+n = 0                           ...12lm + 2ln -nm = 0             ...2From 1, we getl=-m-nSubstituting l=-m-n in 2, we get2-m-nm +2-m-nn-mn = 0-2m2-2mn -2mn-2n2-mn = 02m2+2n2+5mn = 0m+2n 2m+n = 0 m =-2n, -n2If m=-2n, then from 1, we get l=n. If m= -n2 , then from 1, we get l=-n2.Thus, the direction ratios of the two lines are proportional to n, -2n, n and -n2, -n2, n, i.e. 1, -2, 1 and -12, -12, 1.Hence, their direction cosines are±16, ±-26, ±16  ±-16, ±-16, ±26

Page No 27.23:

Question 16:

Find the angle between the lines whose direction cosines are given by the equations
(i) m + n = 0 and l2 + m2 − n2 = 0
(ii) 2l − m + 2n = 0 and mn + nl + lm = 0
(iii) l + 2m + 3n = 0 and 3lm − 4ln + mn = 0
(iv) 2l + 2m − n = 0, mn + ln + lm = 0

Answer:

i Given:l + m + n = 0              ...(1) l2 + m2 - n2 = 0         ...(2)From 1, we getm = -l-nSubstituting m=-l-n in 2, we getl2+ -l-n2-n2l2 + l2 + n2 +2ln - n2 = 02l2 + 2ln = 02l l+n = 0 l = 0 , l = -nIf l=0, then by substituting l=0 in 1, we get m =-n.If l=-n, then by  substituting l=-n  in 1, we get m=0.Thus, the direction ratios of the two lines are proportional to 0, -n, n and -n, 0, n or 0, -1, 1 and-1, 0, 1.Vectors parallel to these lines area = 0i^ -j^ +k^ b = -i^+0j^+k^ If θ is the angle between the lines, then θ is also the angle between a and b.Now,cos θ = a.ba b          =10+1+1 1+0+1           =12 θ=π3

  ii Given:2l - m + 2n = 0                ...(1)mn + nl +lm= 0                ...(2)From 1, we get  m = 2l+2nSubstituting m = 2l+2n in 2, we get2l+2nn+nl+l2l+2n=02ln+2n2+nl+2l2+2ln= 02l2+5ln+2n2= 0 l+2n 2l+n= 0l = -2n , -n2If l =-2n, then by substituting l= -2n in 1, we get m =-2n.If l = -n2, then by substituting l = -n2  in 1, we get m=n.Thus, the direction ratios of the two lines are proportional to-2n, -2n, n and -n2, n, n or -2, -2, 1 and -12, 1, 1.Vectors parallel these lines area = -2i^ -2j^ +k^b = -12i^+ j^+k^ If θ is the angle between the lines, then θ is also the angle between a and b.Now,cos θ=a.ba b        =1-2+14+4+1 14+1+1         =0 θ=π2

iii Given:l + 2m + 3n = 0                            ...(1)3lm - 4ln +mn= 0                        ...(2)From 1, we get  l=-2m-3nSubstituting l =-2m-3n in 2, we get3-2m-3nm-4-2m-3nn+mn=0-6m2-9mn+8mn+12n2+mn=012n2-6m2= 0 m2=2n2m = 2n, -2 nIf m=2n, then by substituting m=2n in 1, we get l =n-22 -3.If m= -2 n, then by substituting m=-2 n in 1, we get l=n22-3.Thus, the direction ratios of the two lines are proportional to n-22 -3,2n, n and n22-3, -2 n, n or -22 -3, 2 , 1 and -22 -3, -2, 1.Vectors parallel to these lines area=-22 -3i^ +2 j^ +k^ b=22 -3i^ -2 j^ +k^If θ is the angle between the lines, then θ is also the angle between a and b.Now,cos θ = a.ba b          =-22 -3i^ +2 j^ +k^.22 -3i^ -2 j^ +k^ 8+9+122+2+1 8+9-122+2+1           =-8-9-2+120+122 20-122           =020+122 20-122          =0θ=π2

(iv) The given relations are

2l + 2m − n = 0                   .....(1)

mn + ln + lm = 0                 .....(2)

From (1), we have

n = 2l + 2m

Putting this value of n in (2), we get

m2l+2m+l2l+2m+lm=02lm+2m2+2l2+2lm+lm=02m2+5lm+2l2=02m+lm+2l=02m+l=0 or m+2l=0l=-2m or l=-m2 

When l=-2m, we have

n=2×-2m+2m=-4m+2m=-2m

When l=-m2, we have

n=2×-m2+2m=-m+2m=m

Thus, the direction ratios of two lines are proportional to

-2m,m,-2m and -m2,m,m

Or -2,1,-2 and -1,2,2

So, vectors parallel to these lines are a=-2i^+j^-2k^ and b=-i^+2j^+2k^.

Let θ be the angle between these lines, then θ is also the angle between a and b.

cosθ=a.bab            =-2i^+j^-2k^.-i^+2j^+2k^4+1+41+4+4            =-2×-1+1×2+-2×23×3            =2+2-49            =0θ=π2

Thus, the angle between the two lines whose direction cosines are given by the given relations is π2.



Page No 27.24:

Question 1:

Define direction cosines of a directed line.

Answer:

The direction cosines of a directed line segment are the cosines of the direction angles of the line segment.Let two points A x1, y1, z1 and B x2, y2, z2  define the directed line segment AB. The direction cosines of AB are given bycos α =x2-x1dcos β =y2-y1dcosγ = z2-z1dHere, d is the distance between A and B.

Page No 27.24:

Question 2:

What are the direction cosines of X-axis?

Answer:

The x-axis makes angles 0°, 90° and 90°with x, y and z axes, respectively.Therefore, the direction cosines of x-axis are cos 0°, cos 90°, cos 90°, i.e. 1, 0, 0.

Page No 27.24:

Question 3:

What are the direction cosines of Y-axis?

Answer:

The y-axis makes angles 90°, 0°and 90° with x, y and z axes, respectively .Therefore, the direction cosines of y-axis are cos 90°, cos 0°, cos 90°, i.e. 0, 1, 0.

Page No 27.24:

Question 4:

What are the direction cosines of Z-axis?

Answer:

The z-axis makes angles 90°, 90° and 0° with x, y and z axes, respectively.Therefore, the direction cosines of z-axis are cos 90°, cos 90°, cos 0°, i.e. 0, 0, 1.

Page No 27.24:

Question 5:

Write the distances of the point (7, −2, 3) from XY, YZ and XZ-planes.

Answer:

The distance of a general point P x, y, z from XY-plane is z.Thus, distance of 7, -2, 3 from XY-plane is 3.Similarly, the distance of P x, y, z from YZ-plane is x.Thus, distance of 7, -2, 3 from YZ- plane is 7.The distance of P x, y, z from XZ-plane is y.Thus, distance of 7, -2, 3 from XZ-plane is 2.

Page No 27.24:

Question 6:

Write the distance of the point (3, −5, 12) from X-axis?

Answer:

The distance of a general point x, y, z from x-axis is y2+z2. Distance of the point 3, -5, 12 from x-axis=-52+122                                                                           = 169                                                                           = 13 units

Page No 27.24:

Question 7:

Write the ratio in which YZ-plane divides the segment joining P (−2, 5, 9) and Q (3, −2, 4).

Answer:

Let the YZ-plane divide the line segment joining points P-2, 5, 9 and Q 3, -2, 4 in the ratio k:1.Using the section formula, the coordinates of the point of intersection are given byk3-2k+1, k-2+5k+1, k4+9k+1On the YZ-plane, the X-coordinate of any point is zero. k3-2k+1=03k-2=0k=23Thus, the YZ-plane divides the line segment formed by joining the given points in the ratio 2:3 internally.

Page No 27.24:

Question 8:

A line makes an angle of 60° with each of X-axis and Y-axis. Find the acute angle made by the line with Z-axis.

Answer:

It is given that a line makes an angle of 60° with both x-axis and y-axis.Suppose the line makes an angle of α with the z-axis.l= cos 60°= 12m= cos 60°=12 n= cos αWe know l2+ m2+n2=1122 +122 + cos α2 =1 14 + 14 + cos 2 α =1cos α=12α=45°Thus, the line makes an angle of 45° with the z-axis.   



Page No 27.25:

Question 9:

If a line makes angles α, β and γ with the coordinate axes, find the value of cos 2α + cos 2β + cos 2γ.

Answer:

It is given that the the line makes angles α, β, γ with the coordinate axis. l=cos α, m=cos β and n=cos γl2+m2+n2=1 cos2 α+cos2 β+cos2 γ=1     ...1Now,cos 2α + cos 2β + cos 2γ = 2cos2 α-1+2cos2 β-1+2cos2 γ-1                                                =2cos2 α+cos2 β+cos2 γ -3                                                =21-3                                                       From 1                                                =-1

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Question 10:

Write the ratio in which the line segment joining (a, b, c) and (−a, −c, −b) is divided by the xy-plane.

Answer:

Suppose the line segment joining the points a, b, c and -a, -c, -b is divided by the XY-plane at a point R in the ratio λ:1. Coordinates of R areλ-a+1aλ+1, λ-c+1bλ+1, λ-b+1cλ+1Since R lies on XY-plane, Z-coordinate of R must be zero.λ-b+1cλ+1 = 0= cb Thus, the required ratio is cb:1 or c:b.Hence, the XY-plane divides the line in the ratio c:b.

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Question 11:

Write the inclination of a line with Z-axis, if its direction ratios are proportional to 0, 1, −1.

Answer:

We know that if a line has direction ratio a, b, c, then the cosine of its angle with the z-axis is given bycos γ = ca2+b2+c2Suppose the inclination of the line with direction ratio 0, 1, -1 with z-axis is γ.Now,cos λ = -10+1+1          =-12 λ=3π4Hence, the inclination of the line with z-axis is 3π4.

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Question 12:

Write the angle between the lines whose direction ratios are proportional to 1, −2, 1 and 4, 3, 2.

Answer:

The direction ratios of the first line are 1, -2, 1 and the direction ratios of the second line are 4, 3, 2.Let θ be the angle between these two lines.Now,cos θ=14+-23+1212+-22+12 42+32+22         =4-6+21+4+116+9+4        =0629       =0 θ=π2Hence, the required angle is  π2.

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Question 13:

Write the distance of the point P (x, y, z) from XOY plane.

Answer:

The distance of the point P x, y, z from the XOY plane is z. 

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Question 14:

Write the coordinates of the projection of point P (x, y, z) on XOZ-plane.

Answer:

The projection of the point P (x, y, z) on XOZ-plane is (x, 0, z) as Y-coordinates of any point on XOZ-plane are equal to zero.

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Question 15:

Write the coordinates of the projection of the point P (2, −3, 5) on Y-axis.

Answer:

The coordinates of the projection of the point P ( 2, -3, 5) on the y-axis are ( 0, -3, 0) as both X and Z coordinates of each point on the y-axis are equal to zero.

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Question 16:

Find the distance of the point (2, 3, 4) from the x-axis.

Answer:

A general point (x, y, z) is at a distance of y2+z2 from the x-axis. Distace of the point (2, 3, 4) from x-axis = 32+42 = 25 = 5 units

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Question 17:

If a line has direction ratios proportional to 2, −1, −2, then what are its direction consines?

Answer:

If a line has direction ratios proportional to 2, -1, and -2, then its direction cosines are     222+-1+-22, -122+-1+-22, -222+-1+-22 = 23, -13, -23Thus, the direction cosines are 23, -13, -23.

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Question 18:

Write direction cosines of a line parallel to z-axis.

Answer:

A line parallel to z-axis, makes an angle of 90°, 90° and 0° with the x, y and z axes, respectively.Thus, the direction cosines are given byl = cos 90° =0 m = cos 90° = 0 n = cos 0 =1Therefore, direction cosines of a line parallel to the z-axis are 0, 0, 1.

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Question 19:

If a unit vector a makes an angle π3 with i^, π4 with j^ and an acute angle θ with k^, then find the value of θ.

Answer:

Since a unit vector makes an angle of π3 with i^, π4 with j^ and an acute angle θ with k^,  l = cos π3 or 12, m =cos π4or12 and n= cos θ.We knowl2+m2+n2=1 122 + 122 + cos2 θ = 114 + 12 + cos2 θ = 1  cos2 θ = 14 cos θ = 12   θ = π3Thus, the vector a makes an angle of π3 with k^. 

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Question 20:

Answer each of the following questions in one word or one sentence or as per exact requirement of the question:

Write the distance of a point P(a, b, c) from x-axis.

Answer:


We know that a general point (x, y, z) has distance y2+z2 from the x-axis.

∴ Distance of a point P(a, b, c) from x-axis = b2+c2

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Question 21:

If a line makes angles 90° and 60° respectively with the positive directions of x and y axes, find the angle which it makes with the positive direction of z-axis.

Answer:

Let the direction cosines of the line be l, m and n.
We know that
l2 + m2 + n2 = 1.
Let the line make angle θ with the positive direction of the z-axis.

α=90°, β=60°, γ=θSo, cos290°+cos260°+cos2θ=10+122+cos2θ=1cos2θ=1-14cos2θ=34cosθ=±32θ=30°or 150° 


cos2α+cos2β+cos2γ=1Here α=60 and β=45 and γ= θSo cos260+cos245+cos2θ=1cos2θ=11412=14cosθ=±12So θ= 60 degree or 120.and here it is given that we have to find the angle made by negative z axisSo cosθ=12θ=120 degree

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Question 1:

For every point P (x, y, z) on the xy-plane,
(a) x = 0
(b) y = 0
(c) z = 0
(d) x = y = z = 0

Answer:

(c) z = 0
            
The Z-coordinate of every point on the XY-plane is zero.

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Question 2:

For every point P (x, y, z) on the x-axis (except the origin),
(a) x = 0, y = 0, z ≠ 0
(b) x = 0, z = 0, y ≠ 0
(c) y = 0, z = 0, x ≠ 0
(d) x = y = z = 0

Answer:

c  y=0, z = 0, x  0              Both Y and Z coordinates on each point of the x-axis are equal to zero. The X-coordinate on the origin is also equal to zero.Therefore, the Y and Z coordinates on each point of the x-axis, except the origin, are equal to zero, while the X-coordinate is non-zero.

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Question 3:

A rectangular parallelopiped is formed by planes drawn through the points (5, 7, 9) and (2, 3, 7) parallel to the coordinate planes. The length of an edge of this rectangular parallelopiped is
(a) 2
(b) 3
(c) 4
(d) all of these

Answer:

d all of theseThe given points 5, 7, 9 and 2, 3, 7 are two diagonally opposite vertices of the parallelopiped as all of their coordinates are different. Edges of the parallelopiped = 5-2, 7-3, 9-7                                                      =3, 4, 2

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Question 4:

A parallelopiped is formed by planes drawn through the points (2, 3, 5) and (5, 9, 7), parallel to the coordinate planes. The length of a diagonal of the parallelopiped is
(a) 7
(b) 38
(c) 155
(d) none of these

Answer:

a 7The given points 2, 3, 5 and 5, 9, 7 are two diagonally opposite vertices of the parallelopiped as all of their coordinates are different.  Edges of the parallelopiped=2-5, 3-9 and 5-7                                                      =3, 6 and 2Now,Length of the diagonal of the parallelopiped=32+62+22                                                             =9+36+4                                                             =49                                                               = 7Hence, length of the diagonal of the parallelopiped formed by the planes parallel to coordinate planes and drawn through points 2, 3, 5 and 5, 9, 7 is 7 units.

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Question 5:

The xy-plane divides the line joining the points (−1, 3, 4) and (2, −5, 6)
(a) internally in the ratio 2 : 3
(b) externally in the ratio 2 : 3
(c) internally in the ratio 3 : 2
(d) externally in the ratio 3 : 2

Answer:

b externally in the ratio 2:3Let the XY-plane divide the line segment joining points P-1, 3, 4 and Q2, -5, 6 in the ratio k:1.Using the section formula, the coordinates of the point of intersection are given byk2-1k+1, k-5+3k+1, k6+4k+1On the XY-plane, the Z-coordinate of any point is zero. k6+4k+1=06k+4=0k=-23Thus, the  XY-plane divides the line segment joining the given points in the ratio 2:3 externally.

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Question 6:

If the x-coordinate of a point P on the join of Q (2, 2, 1) and R (5, 1, −2) is 4, then its z-coordinate is
(a) 2
(b) 1
(c) −1
(d) −2

Answer:

c -1Suppose the point P divides the line joining the point Q 2, 2, 1 and R 5, 1, -2 in the ratio k:1.Using the section formula, the coordinates of the point of intersection are given byk5+2k+1, k1+2k+1, k-2+1k+1It is given that the X-coordinate of P is 4.k5+2k+1=45k+2=4k+1 k=2Now,Z-coordinate of P =k-2+1k+1                                  = 2-2+12+1            Substituting k=2                                  =-1                                                            

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Question 7:

The distance of the point P (a, b, c) from the x-axis is
(a) b2+c2

(b) a2+c2

(c) a2+b2

(d) none of these

Answer:

a b2+c2The projection of the point P a, b, c on the x-axis is a, 0, 0 as both Y and Z coordinates on any point on the x-axis are equal to zero.  Distance of P a, b, c from x-axis = Distance of P a, b, c from a, 0, 0                                                                   = a-a2+b-02+c-02                                                                   =b2+c2



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Question 8:

Ratio in which the xy-plane divides the join of (1, 2, 3) and (4, 2, 1) is
(a) 3 : 1 internally
(b) 3 : 1 externally
(c) 1 : 2 internally
(d) 2 : 1 externally

Answer:

b 3:1 externallySuppose the XY-plane divides the line segment joining the points P 1, 2, 3 and Q 4, 2, 1 in the ratio k:1.Using the section formula, the coordinates of the point of intersection are given byk4+1k+1, k2+2k+1,k1+3k+1The Z-coordinate of any point on the XY-plane is zero.k1+3k+1=0k+3 = 0k=-3=-31Thus, the XY-plane divides the line segment joining the given points in the ratio 3:1 externally.

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Question 9:

If P (3, 2, −4), Q (5, 4, −6) and R (9, 8, −10) are collinear, then R divides PQ in the ratio
(a) 3 : 2 internally
(b) 3 : 2 externally
(c) 2 : 1 internally
(d) 2 : 1 externally

Answer:

b 3:2 externallySuppose the point R divides PQ in the ratio λ:1.Coordinates of R are 5λ+3λ+1, 4λ+2λ+1, -6λ-4λ+1.But the coordinates of R are 9, 8, -10. 5λ+3λ+1=9, 4λ+2λ+1=8 and -6λ-4λ+1=-10From each of these equations, we getλ=-32 R divides PQ in the ratio 3:2 externally.

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Question 10:

A (3, 2, 0), B (5, 3, 2) and C (−9, 6, −3) are the vertices of a triangle ABC. If the bisector of ∠ABC meets BC at D, then coordinates of D are
(a) (19/8, 57/16, 17/16)
(b) (−19/8, 57/16, 17/16)
(c) (19/8, −57/16, 17/16)
(d) none of these

Answer:

Since the bisector of ABC cannot meet BC, the solution of this question is not possible.

Disclaimer:This question is wrong, so the solution has not been provide.

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Question 11:

If O is the origin, OP = 3 with direction ratios proportional to −1, 2, −2 then the coordinates of P are
(a) (−1, 2, −2)
(b) (1, 2, 2)
(c) (−1/9, 2/9, −2/9)
(d) (3, 6, −9)

Answer:

a -1, 2,-2Let the coordinates of P be x, y, z. Then,Direction ratios of OP= Coordinates of P- Coordinates of O-1, 2, 2 = x-0, y-0, z-0Thus, coordinates of P are -1, 2,-2.

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Question 12:

The angle between the two diagonals of a cube is
(a) 30°

(b) 45°

(c) cos-113

(d) cos-113

Answer:

(d) cos-113



Let a be the length of an edge of the cube and let one corner be at the origin as shown in the figure. Clearly, OP, AR, BS, and CQ are the diagonals of the cube.Consider the diagonals OP and AR.Direction ratios of OP and AR are proportional to a-0, a-0, a-0 and 0-a, a-0, a-0, i.e. a, a, a and -a, a, a, respectively.Let θ be the angle between OP and AR. Then,cos θ =a×-a+a×a+a×aa2+a2+a2-a2+a2+a2cos θ =-a2+a2+a23a23a2 cos θ =13 θ = cos-1 13 Similarly, the angles between other pairs of the diagonals are equal to cos-1 13 as the angle between any two diagonals of a cube is cos-1 13.

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Question 13:

If a line makes angles α, β, γ, δ with four diagonals of a cube, then cos2 α + cos2 β + cos2 γ + cos2 δ is equal to

(a) 13

(b) 23

(c) 43

(d) 83

Answer:

c 43




Let a be the length of an edge of the cube and let one corner be at the origin as shown in the figure. Clearly, OP, AR, BS and CQ are the diagonals of the cube. The direction ratios of OP, AR, BS and CQ area-0, a-0, a-0, i.e. a, a, a0-a, a-0, a-0, i.e. -a, a, aa-0, 0-a, a-0, i.e. a, -a, aa-0, a-0, 0-a, i.e. a, a,-a Let the direction ratios of a line be proportional to l, m and n. Suppose this line makes angles α, β, γ and δ with OP, AR, BS and CQ, respectively.Now, α is the angle between OP and the line whose direction ratios are proportional to l, m and n.                cos α = a.l+a.m+a.na2+a2+a2l2+m2+n2cos α = l + m + n3l2+m2+n2Since β is the angle between AR and the line with direction ratios proportional to l, m and n, we get               cos β = -a.l+a.m+a.na2+a2+a2l2+m2+n2 cos β=-l+m+n3l2+m2+n2Similarly,               cos γ = a.l-a.m+a.na2+a2+a2l2+m2+n2 cos γ=l - m + n3l2+m2+n2               cos δ = a.l+a.m-a.na2+a2+a2l2+m2+n2 cos δ=l+m - n3l2+m2+n2cos2α + cos2β + cos2γ+cos2δ                  =l+m+n23l2+m2+n2+-l+m+n23l2+m2+n2+I-m+n23l2+m2+n2+l+m - n23l2+m2+n2                 =13l2+m2+n2l+m+n2+-l+m+n2+I-m+n2+l+m -n2                 =13l2+m2+n24l2+m2+n2 =43


 



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