Rd Sharma XII Vol 2 2021 Solutions for Class 12 Commerce Maths Chapter 10 The Plane are provided here with simple step-by-step explanations. These solutions for The Plane are extremely popular among Class 12 Commerce students for Maths The Plane Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma XII Vol 2 2021 Book of Class 12 Commerce Maths Chapter 10 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma XII Vol 2 2021 Solutions. All Rd Sharma XII Vol 2 2021 Solutions for class Class 12 Commerce Maths are prepared by experts and are 100% accurate.

#### Question 1:

Find the vector equation of a plane passing through a point with position vector $2\stackrel{^}{i}-\stackrel{^}{j}+\stackrel{^}{k}$ and perpendicular to the vector $4\stackrel{^}{i}+2\stackrel{^}{j}-3\stackrel{^}{k}.$

#### Question 2:

Find the Cartesian form of the equation of a plane whose vector equation is
(i) $\stackrel{\to }{r}·\left(12\stackrel{^}{i}-3\stackrel{^}{j}+4\stackrel{^}{k}\right)+5=0$

(ii) $\stackrel{\to }{r}·\left(-\stackrel{^}{i}+\stackrel{^}{j}+2\stackrel{^}{k}\right)=9$

#### Question 3:

Find the vector equations of the coordinate planes.

#### Question 4:

Find the vector equation of each one of following planes.
(i) 2xy + 2z = 8
(ii) x + yz = 5
(iii) x + y = 3

#### Question 5:

Find the vector and Cartesian equations of a plane passing through the point (1, −1, 1) and normal to the line joining the points (1, 2, 5) and (−1, 3, 1).

#### Question 6:

is a vector of magnitude $\sqrt{3}$ and is equally inclined to an acute angle with the coordinate axes. Find the vector and Cartesian forms of the equation of a plane which passes through (2, 1, −1) and is normal to .

#### Question 7:

The coordinates of the foot of the perpendicular drawn from the origin to a plane are (12, −4, 3). Find the equation of the plane.

#### Question 8:

Find the equation of the plane passing through the point (2, 3, 1), given that the direction ratios of the normal to the plane are proportional to 5, 3, 2.

#### Question 9:

If the axes are rectangular and P is the point (2, 3, −1), find the equation of the plane through P at right angles to OP.

#### Question 10:

Find the intercepts made on the coordinate axes by the plane 2x + y − 2z = 3 and also find the direction cosines of the normal to the plane.

#### Question 11:

A plane passes through the point (1, −2, 5) and is perpendicular to the line joining the origin to the point $3\stackrel{^}{i}+\stackrel{^}{j}-\stackrel{^}{k}.$ Find the vector and Cartesian forms of the equation of the plane.

#### Question 12:

Find the equation of the plane that bisects the line segment joining the points (1, 2, 3) and (3, 4, 5) and is at right angle to it.

#### Question 13:

Show that the normals to the following pairs of planes are perpendicular to each other.

(i) xy + z − 2 = 0 and 3x + 2yz + 4 = 0

(ii)

#### Question 14:

Show that the normal vector to the plane 2x + 2y + 2z = 3 is equally inclined to the coordinate axes.

#### Question 15:

Find a vector of magnitude 26 units normal to the plane 12x − 3y + 4z = 1.

#### Question 16:

If the line drawn from (4, −1, 2) meets a plane at right angles at the point (−10, 5, 4), find the equation of the plane.

#### Question 17:

Find the equation of the plane which bisects the line segment joining the points (−1, 2, 3) and (3, −5, 6) at right angles.

#### Question 18:

Find the vector and Cartesian equations of the plane that passes through the point (5, 2, −4) and is perpendicular to the line with direction ratios 2, 3, −1.

#### Question 19:

If O be the origin and the coordinates of P be (1, 2,−3), then find the equation of the plane passing through P and perpendicular to OP.

#### Question 20:

If O is the origin and the coordinates of A are (abc). Find the direction cosines of OA and the equation of the plane through A at right angles to OA.              [NCERT EXEMPLAR]

It is given that O is the origin and the coordinates of A are (abc).

The direction ratios of OA are proportional to

$a-0,b-0,c-0$ or a, b, c

∴ Direction cosines of OA are

$\frac{a}{\sqrt{{a}^{2}+{b}^{2}+{c}^{2}}},\frac{b}{\sqrt{{a}^{2}+{b}^{2}+{c}^{2}}},\frac{c}{\sqrt{{a}^{2}+{b}^{2}+{c}^{2}}}$

The normal vector to the required plane is $a\stackrel{^}{i}+b\stackrel{^}{j}+c\stackrel{^}{k}$.

The vector equation of the plane through A(abc) and perpendicular to OA is

The Cartesian equation of this plane is

#### Question 21:

Find the vector equation of the plane with intercepts 3, –4 and 2 on x, y and z-axis respectively.

The equation of the plane in the intercept form is $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$, where ab and are the intercepts on the xy and z-axis, respectively.

It is given that the intercepts made by the plane on the xy and z-axis are 3, –4 and 2, respectively.

∴ a = 3, b = −4, c = 2

Thus, the equation of the plane is

$\frac{x}{3}+\frac{y}{\left(-4\right)}+\frac{z}{2}=1\phantom{\rule{0ex}{0ex}}⇒4x-3y+6z=12$
$⇒\left(x\stackrel{^}{i}+y\stackrel{^}{j}+z\stackrel{^}{k}\right).\left(4\stackrel{^}{i}-3\stackrel{^}{j}+6\stackrel{^}{k}\right)=12\phantom{\rule{0ex}{0ex}}⇒\stackrel{\to }{r}.\left(4\stackrel{^}{i}-3\stackrel{^}{j}+6\stackrel{^}{k}\right)=12$
This is the vector form of the equation of the given plane.

#### Question 1:

Find the vector equation of a plane which is at a distance of 3 units from the origin and has $\stackrel{^}{k}$ as the unit vector normal to it.

#### Question 2:

Find the vector equation of a plane which is at a distance of 5 units from the origin and which is normal to the vector $\stackrel{^}{i}-2\stackrel{^}{j}-2\stackrel{^}{k}.$

#### Question 3:

Reduce the equation 2x − 3y − 6z = 14 to the normal form and, hence, find the length of the perpendicular from the origin to the plane. Also, find the direction cosines of the normal to the plane.

#### Question 4:

Reduce the equation $\stackrel{\to }{r}·\left(\stackrel{^}{i}-2\stackrel{^}{j}+2\stackrel{^}{k}\right)+6=0$ to normal form and, hence, find the length of the perpendicular from the origin to the plane.

#### Question 5:

Write the normal form of the equation of the plane 2x − 3y + 6z + 14 = 0.

#### Question 6:

The direction ratios of the perpendicular from the origin to a plane are 12, −3, 4 and the length of the perpendicular is 5. Find the equation of the plane.

#### Question 7:

Find a unit normal vector to the plane x + 2y + 3z − 6 = 0.

#### Question 8:

Find the equation of a plane which is at a distance of $3\sqrt{3}$ units from the origin and the normal to which is equally inclined to the coordinate axes.

#### Question 9:

Find the equation of the plane passing through the point (1, 2, 1) and perpendicular to the line joining the points (1, 4, 2) and (2, 3, 5). Find also the perpendicular distance of the origin from this plane.

#### Question 10:

Find the vector equation of the plane which is at a distance of $\frac{6}{\sqrt{29}}$ from the origin and its normal vector from the origin is $2\stackrel{^}{i}-3\stackrel{^}{j}+4\stackrel{^}{k}.$ Also, find its Cartesian form.

#### Question 11:

Find the distance of the plane 2x − 3y + 4z − 6 = 0 from the origin.

#### Question 1:

Find the vector equation of the plane passing through the points (1, 1, 1), (1, −1, 1) and (−7, −3, −5).

#### Question 2:

Find the vector equation of the plane passing through the points P (2, 5, −3), Q (−2, −3, 5) and R (5, 3, −3).

#### Question 3:

Find the vector equation of the plane passing through points A (a, 0, 0), B (0, b, 0) and C (0, 0, c). Reduce it to normal form. If plane ABC is at a distance p from the origin, prove that $\frac{1}{{p}^{2}}=\frac{1}{{a}^{2}}+\frac{1}{{b}^{2}}+\frac{1}{{c}^{2}}.$

#### Question 4:

Find the vector equation of the plane passing through the points (1, 1, −1), (6, 4, −5) and (−4, −2, 3).

#### Question 5:

Find the vector equation of the plane passing through the points

#### Question 1:

Find the angle between the given planes.
(i)

(ii)

(iii)

#### Question 2:

Find the angle between the planes.
(i) 2xy + z = 4 and x + y + 2z = 3
(ii) x + y − 2z = 3 and 2x − 2y + z = 5
(iii) xy + z = 5 and x + 2y + z = 9
(iv) 2x − 3y + 4z = 1 and − x + y = 4
(v) 2x + y − 2z = 5 and 3x − 6y − 2z = 7

#### Question 3:

Show that the following planes are at right angles.
(i)

(ii) x − 2y + 4z = 10 and 18x + 17y + 4z = 49

#### Question 4:

Determine the value of λ for which the following planes are perpendicular to each other.
(i)

(ii) 2x − 4y + 3z = 5 and x + 2y + λz = 5

(iii) 3x − 6y − 2z = 7 and 2x + y − λz = 5

#### Question 5:

Find the equation of a plane passing through the point (−1, −1, 2) and perpendicular to the planes 3x + 2y − 3z = 1 and 5x − 4y + z = 5.

#### Question 6:

Obtain the equation of the plane passing through the point (1, −3, −2) and perpendicular to the planes x + 2y + 2z = 5 and 3x + 3y + 2z = 8.

#### Question 7:

Find the equation of the plane passing through the origin and perpendicular to each of the planes x + 2yz = 1 and 3x − 4y + z = 5.

#### Question 8:

Find the equation of the plane passing through the points (1, −1, 2) and (2, −2, 2) and which is perpendicular to the plane 6x − 2y + 2z = 9.

#### Question 9:

Find the equation of the plane passing through the points (2, 2, 1) and (9, 3, 6) and perpendicular to the plane 2x + 6y + 6z = 1.

#### Question 10:

Find the equation of the plane passing through the points whose coordinates are (−1, 1, 1) and (1, −1, 1) and perpendicular to the plane x + 2y + 2z = 5.

#### Question 11:

Find the equation of the plane with intercept 3 on the y-axis and parallel to the ZOX plane.

#### Question 12:

Find the equation of the plane that contains the point (1, −1, 2) and is perpendicular to each of the planes 2x + 3y − 2z = 5 and x + 2y − 3z = 8.

#### Question 13:

Find the equation of the plane passing through (a, b, c) and parallel to the plane $\stackrel{\to }{r}·\left(\stackrel{^}{i}+\stackrel{^}{j}+\stackrel{^}{k}\right)=2.$

#### Question 14:

Find the equation of the plane passing through the point (−1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0.

#### Question 15:

Find the vector equation of the plane through the points (2, 1, −1) and (−1, 3, 4) and perpendicular to the plane x − 2y + 4z = 10.

#### Question 1:

Find the vector equations of the following planes in scalar product form $\left(\stackrel{\to }{r}·\stackrel{\to }{n}=d\right):$
(i) $\stackrel{\to }{r}=\left(2\stackrel{^}{i}-\stackrel{^}{k}\right)+\lambda \stackrel{^}{i}+\mu \left(\stackrel{^}{i}-2\stackrel{^}{j}-\stackrel{^}{k}\right)$

(ii)

(iii) $\stackrel{\to }{r}=\left(\stackrel{^}{i}+\stackrel{^}{j}\right)+\lambda \left(\stackrel{^}{i}+2\stackrel{^}{j}-\stackrel{^}{k}\right)+\mu \left(-\stackrel{^}{i}+\stackrel{^}{j}-2\stackrel{^}{k}\right)$

(iv) $\stackrel{\to }{r}=\stackrel{^}{i}-\stackrel{^}{j}+\lambda \left(\stackrel{^}{i}+\stackrel{^}{j}+\stackrel{^}{k}\right)+\mu \left(4\stackrel{^}{i}-2\stackrel{^}{j}+3\stackrel{^}{k}\right)$

Disclaimer: The answer given for part (iv) of this problem in the text book is incorrect.

#### Question 2:

Find the Cartesian forms of the equations of the following planes.
(i) $\stackrel{\to }{r}=\left(\stackrel{^}{i}-\stackrel{^}{j}\right)+s\left(-\stackrel{^}{i}+\stackrel{^}{j}+2\stackrel{^}{k}\right)+t\left(\stackrel{^}{i}+2\stackrel{^}{j}+\stackrel{^}{k}\right)$

(ii) $\stackrel{\to }{r}=\left(1+s+t\right)\stackrel{^}{i}+\left(2-s+t\right)\stackrel{^}{i}+\left(3-2s+2t\right)\stackrel{^}{k}$

#### Question 3:

Find the vector equation of the following planes in non-parametric form.
(i)

(ii) $\stackrel{\to }{r}=\left(2\stackrel{^}{i}+2\stackrel{^}{j}-\stackrel{^}{k}\right)+\lambda \left(\stackrel{^}{i}+2\stackrel{^}{j}+3\stackrel{^}{k}\right)+\mu \left(5\stackrel{^}{i}-2\stackrel{^}{j}+7\stackrel{^}{k}\right)$

#### Question 1:

Find the equation of the plane which is parallel to 2x − 3y + z = 0 and which passes through (1, −1, 2).

#### Question 2:

Find the equation of the plane through (3, 4, −1) which is parallel to the plane $\stackrel{\to }{r}·\left(2\stackrel{^}{i}-3\stackrel{^}{j}+5\stackrel{^}{k}\right)+2=0.$

#### Question 3:

Find the equation of the plane passing through the line of intersection of the planes 2x − 7y + 4z − 3 = 0, 3x − 5y + 4z + 11 = 0 and the point (−2, 1, 3).

#### Question 4:

Find the equation of the plane through the point $2\stackrel{^}{i}+\stackrel{^}{j}-\stackrel{^}{k}$ and passing through the line of intersection of the planes

#### Question 5:

Find the equation of the plane passing through the line of intersection of the planes 2xy = 0 and 3zy = 0 and perpendicular to the plane 4x + 5y − 3z = 8.

#### Question 6:

Find the equation of the plane which contains the line of intersection of the planes x + 2y + 3z − 4 = 0 and 2x + yz + 5 = 0 and which is perpendicular to the plane 5x + 3y − 6z + 8 = 0.

#### Question 7:

Find the equation of the plane through the line of intersection of the planes x + 2y + 3z + 4 = 0 and xy + z + 3 = 0 and passing through the origin.

#### Question 8:

Find the vector equation (in scalar product form) of the plane containing the line of intersection of the planes x − 3y + 2z − 5 = 0 and 2xy + 3z − 1 = 0 and passing through (1, −2, 3).

#### Question 9:

Find the equation of the plane that is perpendicular to the plane 5x + 3y + 6z + 8 = 0 and which contains the line of intersection of the planes x + 2y + 3z − 4 = 0, 2x + yz + 5 = 0.

#### Question 10:

Find the equation of the plane through the line of intersection of the planes which is at a unit distance from the origin.

#### Question 11:

Find the equation of the plane passing through the intersection of the planes 2x + 3yz + 1 = 0 and x + y − 2z + 3 = 0 and perpendicular to the plane 3xy − 2z − 4 = 0.

#### Question 12:

Find the equation of the plane that contains the line of intersection of the planes and which is perpendicular to the plane $\stackrel{\to }{r}·\left(5\stackrel{^}{i}+3\stackrel{^}{j}-6\stackrel{^}{k}\right)+8=0.$

#### Question 13:

Find the vector equation of the plane passing through the intersection of the planes  and the point (1, 1, 1).

Equation plane passing through intersection of
$\stackrel{\to }{r}·\left(\stackrel{^}{i}+\stackrel{^}{j}+\stackrel{^}{k}=6\right)$ and $\stackrel{\to }{r}·\left(2\stackrel{^}{i}+3\stackrel{^}{j}+4\stackrel{^}{k}\right)=-5$ is
$\left[\stackrel{\to }{r}·\left(\stackrel{^}{i}+\stackrel{^}{j}+\stackrel{^}{k}\right)-6\right]+\lambda \left[\stackrel{\to }{r}·\left(2\stackrel{^}{i}+3\stackrel{^}{j}+4\stackrel{^}{k}\right)+5\right]=0$
$⇒\stackrel{\to }{r}·\left[\left(1+2\lambda \right)\stackrel{^}{i}+\left(1+3\lambda \right)\stackrel{^}{j}+\left(1+4\lambda \right)\stackrel{^}{k}\right]-6+5\lambda =0$
Since, above planes passes through (1, 1, 1), then vector $\left(\stackrel{^}{i}+\stackrel{^}{j}+\stackrel{^}{k}\right)$ should satisfy it
$\therefore \left(\stackrel{^}{i}+\stackrel{^}{j}+\stackrel{^}{k}\right)·\left[\left(1+2\lambda \right)\stackrel{^}{i}+\left(1+3\lambda \right)\stackrel{^}{j}+\left(1+4\lambda \right)\stackrel{^}{k}\right]-6+5\lambda =0$
$⇒1+2\lambda +1+3\lambda +1+4\lambda -6+5\lambda =0\phantom{\rule{0ex}{0ex}}⇒-3+14\lambda =0\phantom{\rule{0ex}{0ex}}⇒\lambda =\frac{3}{14}$
Hence, equation of plane will be
$\stackrel{\to }{r}·\left[\left(1+2×\frac{3}{14}\right)\stackrel{^}{i}+\left(1+3×\frac{3}{14}\right)\stackrel{^}{j}+\left(1+4×\frac{3}{14}\right)\stackrel{^}{k}\right]=6-5×\frac{3}{14}\phantom{\rule{0ex}{0ex}}⇒\stackrel{\to }{r}·\left(20\stackrel{^}{i}+23\stackrel{^}{j}+26\stackrel{^}{k}\right)=69$

#### Question 14:

Find the equation of the plane passing through the intersection of the planes and the point (2, 1, 3).

#### Question 1:

Find the equation of the plane passing through the following points.
(i) (2, 1, 0), (3, −2, −2) and (3, 1, 7)
(ii) (−5, 0, −6), (−3, 10, −9) and (−2, 6, −6)
(iii) (1, 1, 1), (1, −1, 2) and (−2, −2, 2)
(iv) (2, 3, 4), (−3, 5, 1) and (4, −1, 2)
(v) (0, −1, 0), (3, 3, 0) and (1, 1, 1)

(i) The equation of the plane passing through points (2, 1, 0), (3, −2, −2) and (3, 1, 7) is given by

(ii) The equation of the plane passing through points (−5, 0, −6), (−3, 10, −9) and (−2, 6, −6) is given by

(iii) The equation of the plane passing through points  (1, 1, 1), (1, −1, 2) and (−2, −2, 2) is given by

(iv) The equation of the plane passing through points (2, 3, 4), (−3, 5, 1) and (4, −1, 2) is given by

(v) The equation of the plane passing through points (0, −1, 0), (3, 3, 0) and (1, 1, 1) is given by

#### Question 15:

Find the equation of the plane through the intersection of the planes 3xy + 2z = 4 and x + y + z = 2 and the point (2, 2, 1).

#### Question 16:

Find the vector equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane xy + z = 0.

#### Question 17:

Find the equation of the plane passing through () and parallel to the plane $\stackrel{\to }{r}·\left(\stackrel{^}{i}+\stackrel{^}{j}+\stackrel{^}{k}\right)=2$.

Given: Plane equation: $\stackrel{\to }{r}.\left(\stackrel{^}{i}+\stackrel{^}{j}+\stackrel{^}{k}\right)=2$
Now, the equation of plane parallel to  is
Since the plane passes through
$\therefore \left(a\stackrel{^}{i}+b\stackrel{^}{j}+c\stackrel{^}{k}\right)·\left(\stackrel{^}{i}+\stackrel{^}{j}+\stackrel{^}{k}\right)=\lambda \phantom{\rule{0ex}{0ex}}⇒a+b+c=\lambda$
Substituting the value of $\lambda$ in above plane equation, we get
$\stackrel{\to }{r}·\left(\stackrel{^}{i}+\stackrel{^}{j}+\stackrel{^}{k}\right)=\left(a+b+c\right)\phantom{\rule{0ex}{0ex}}⇒\left(x\stackrel{^}{i}+y\stackrel{^}{j}+z\stackrel{^}{k}\right)·\left(\stackrel{^}{i}+\stackrel{^}{j}+\stackrel{^}{k}\right)=a+b+c\phantom{\rule{0ex}{0ex}}⇒x+y+z=a+b+c$
Hence, required equation of plane is $x+y+z=a+b+c$

#### Question 18:

Find the equation of the plane which contains the line of intersection of the planes x$+$2y$+$34$=$0 and 2$x+y-z$ $+$ 5$=$0 and whose x-intercept is twice its z-intercept. Hence, write the equation of the plane passing through the point (2, 3, $-$1) and parallel to the plane obtained above.

The equation of the family of planes passing through the intersection of the planes x + 2y + 3z − 4 = 0 and 2xy − z + 5 = 0 is

(x + 2y + 3z − 4) + k(2x + y − z + 5) = 0, where k is some constant

$⇒\left(2k+1\right)x+\left(k+2\right)y+\left(3-k\right)z=4-5k\phantom{\rule{0ex}{0ex}}⇒\frac{x}{\left(\frac{4-5k}{2k+1}\right)}+\frac{y}{\left(\frac{4-5k}{k+2}\right)}+\frac{z}{\left(\frac{4-5k}{3-k}\right)}=1$
It is given that x-intercept of the required plane is twice its z-intercept.

$\left(\frac{4-5k}{2k+1}\right)=2\left(\frac{4-5k}{3-k}\right)\phantom{\rule{0ex}{0ex}}⇒\left(4-5k\right)\left(3-k\right)=\left(4k+2\right)\left(4-5k\right)\phantom{\rule{0ex}{0ex}}⇒\left(4-5k\right)\left(3-k-4k-2\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(4-5k\right)\left(1-5k\right)=0$

When $k=\frac{4}{5}$, the equation of the plane is $\left(2×\frac{4}{5}+1\right)x+\left(\frac{4}{5}+2\right)y+\left(3-\frac{4}{5}\right)z=4-5×\frac{4}{5}⇒13x+14y+11z=0$.

This plane does not satisfies the given condition, so this is rejected.

When $k=\frac{1}{5}$, the equation of the plane is $\left(2×\frac{1}{5}+1\right)x+\left(\frac{1}{5}+2\right)y+\left(3-\frac{1}{5}\right)z=4-5×\frac{1}{5}⇒7x+11y+14z=15$.

Thus, the equation of the required plane is 7x + 11y + 14z = 15.

Also, the equation of the plane passing through the point (2, 3, −1) and parallel to the plane 7x + 11y + 14z = 15 is

$7\left(x-2\right)+11\left(y-3\right)+14\left(z+1\right)=0\phantom{\rule{0ex}{0ex}}⇒7x+11y+14z=33$

#### Question 19:

Find the equation of the plane through the line of intersection of the planes $x+y+z=$1 and 2x$+$3$y+$4$z=$5 and twice of its $y$-intercept is equal to three times its $z$-intercept.

The equation of the family of the planes passing through the intersection of the planes xyz = 1 and 2x + 3y + 4z = 5 is

(x + y + z − 1) + k(2x + 3y + 4z − 5) = 0, where k is some constant

⇒ (2k + 1)x + (3k + 1)y + (4k + 1)z = 5k + 1                .....(1)

It is given that twice of y-intercept is equal to three times its z-intercept.

$\therefore 2\left(\frac{5k+1}{3k+1}\right)=3\left(\frac{5k+1}{4k+1}\right)\phantom{\rule{0ex}{0ex}}⇒\left(5k+1\right)\left(8k+2-9k-3\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(5k+1\right)\left(-k-1\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(5k+1\right)\left(k+1\right)=0$

Putting $k=-\frac{1}{5}$ in (1), we get

$\left(-\frac{2}{5}+1\right)x+\left(-\frac{3}{5}+1\right)y+\left(-\frac{4}{5}+1\right)z=5×\left(-\frac{1}{5}\right)+1\phantom{\rule{0ex}{0ex}}⇒3x+2y+z=0$
This plane passes through the origin. So, the intercepts made by the plane with the coordinate axes is 0. Hence, this equation of plane is not accepted as twice of y-intercept is not equal to three times its z-intercept.

Putting $k=-1$ in (1), we get

$\left(-2+1\right)x+\left(-3+1\right)y+\left(-4+1\right)z=5×\left(-1\right)+1\phantom{\rule{0ex}{0ex}}⇒-x-2y-3z=-4\phantom{\rule{0ex}{0ex}}⇒x+2y+3z=4$
Here, twice of y-intercept is equal to three times its z-intercept.

Thus, the equation of the required plane is x + 2y + 3z = 4.

#### Question 1:

Find the distance of the point $2\stackrel{^}{i}-\stackrel{^}{j}-4\stackrel{^}{k}$ from the plane $\stackrel{\to }{r}·\left(3\stackrel{^}{i}-4\stackrel{^}{j}+12\stackrel{^}{k}\right)-9=0.$

#### Question 2:

Show that the points are equidistant from the plane $\stackrel{\to }{r}·\left(5\stackrel{^}{i}+2\stackrel{^}{j}-7\stackrel{^}{k}\right)+9=0.$

#### Question 3:

Find the distance of the point (2, 3, −5) from the plane x + 2y − 2z − 9 = 0.

#### Question 4:

Find the equations of the planes parallel to the plane x + 2y − 2z + 8 = 0 that are at a distance of 2 units from the point (2, 1, 1).

#### Question 5:

Show that the points (1, 1, 1) and (−3, 0, 1) are equidistant from the plane 3x + 4y − 12z + 13 = 0.

#### Question 6:

Find the equations of the planes parallel to the plane x − 2y + 2z − 3 = 0 and which are at a unit distance from the point (1, 1, 1).

#### Question 7:

Find the distance of the point (2, 3, 5) from the xy - plane.

#### Question 8:

Find the distance of the point (3, 3, 3) from the plane $\stackrel{\to }{r}·\left(5\stackrel{^}{i}+2\stackrel{^}{j}-7k\right)+9=0$

#### Question 9:

If the product of the distances of the point (1, 1, 1) from the origin and the plane xy + z + λ = 0 be 5, find the value of λ.

Disclaimer: The answer or problem given in the text book is incorrect for this.

#### Question 10:

Find an equation for the set of all points that are equidistant from the planes 3x − 4y + 12z = 6 and 4x + 3z = 7.

#### Question 11:

Find the distance between the point (7, 2, 4) and the plane determined by the points A(2, 5, −3), B(−2, −3, 5) and C(5, 3, −3).         [CBSE 2014]

The given points are A(2, 5, −3), B(−2, −3, 5) and C(5, 3, −3).

The equation of the plane ABC is given by

$\left|\begin{array}{ccc}x-{x}_{1}& y-{y}_{1}& z-{z}_{1}\\ {x}_{2}-{x}_{1}& {y}_{2}-{y}_{1}& {z}_{2}-{z}_{1}\\ {x}_{3}-{x}_{1}& {y}_{3}-{y}_{1}& {z}_{3}-{z}_{1}\end{array}\right|=0\phantom{\rule{0ex}{0ex}}⇒\left|\begin{array}{ccc}x-2& y-5& z-\left(-3\right)\\ -2-2& -3-5& 5-\left(-3\right)\\ 5-2& 3-5& -3-\left(-3\right)\end{array}\right|=0\phantom{\rule{0ex}{0ex}}⇒\left|\begin{array}{ccc}x-2& y-5& z+3\\ -4& -8& 8\\ 3& -2& 0\end{array}\right|=0\phantom{\rule{0ex}{0ex}}⇒\left|\begin{array}{ccc}x-2& y-5& z+3\\ 1& 2& -2\\ 3& -2& 0\end{array}\right|=0\phantom{\rule{0ex}{0ex}}⇒-4\left(x-2\right)-6\left(y-5\right)-8\left(z+3\right)=0\phantom{\rule{0ex}{0ex}}⇒2\left(x-2\right)+3\left(y-5\right)+4\left(z+3\right)=0\phantom{\rule{0ex}{0ex}}⇒2x+3y+4z-7=0$

∴ Distance between the point (7, 2, 4) and the plane $2x+3y+4z-7=0$

= Length of perpendicular from (7, 2, 4) to the plane $2x+3y+4z-7=0$

Thus, the required distance between the given point and the plane is $\sqrt{29}$ units.

#### Question 12:

A plane makes intercepts −6, 3, 4 respectively on the coordinate axes. Find the length of the perpendicular from the origin on it.       [CBSE 2014]

We know that the equation of the plane which makes intercepts a, b and c on the coordinate axes is $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$.

So, the equation of the plane which makes intercepts −6, 3, 4 on the x-axis, y-axis and z-axis, respectively is

$\frac{x}{-6}+\frac{y}{3}+\frac{z}{4}=1\phantom{\rule{0ex}{0ex}}⇒-2x+4y+3z=12\phantom{\rule{0ex}{0ex}}⇒2x-4y-3z+12=0$

∴ Length of the perpendicular from (0, 0, 0) to the plane $2x-4y-3z+12=0$

Thus, the length of the perpendicular from the origin to the plane is $\frac{12}{\sqrt{29}}$ units.

#### Question 13:

Find the distance of the point (1, $-$2, 4) from plane passing throuhg the point (1, 2, 2) and perpendicular of the planes $x-y+$2$z=$3 and 2$x-$2$y+z+$12$=$0.

Let the equation of plane passing through the point (1, 2, 2) be

$a\left(x-1\right)+b\left(y-2\right)+c\left(z-2\right)=0$          .....(1)

Here, abc are the direction ratios of the normal to the plane.

The equations of the given planes are x − y + 2z = 3 and 2x − 2y + z + 12 = 0.

Plane (1) is perpendicular to the given planes.

a − b + 2c = 0           .....(2)

2a − 2b + c = 0         .....(3)

Eliminating ab and c from (1), (2) and (3), we get

$\left|\begin{array}{ccc}x-1& y-2& z-2\\ 1& -1& 2\\ 2& -2& 1\end{array}\right|=0\phantom{\rule{0ex}{0ex}}⇒\left(x-1\right)\left(-1+4\right)-\left(y-2\right)\left(1-4\right)+\left(z-2\right)\left(-2+2\right)=0\phantom{\rule{0ex}{0ex}}⇒3x+3y-9=0\phantom{\rule{0ex}{0ex}}⇒x+y-3=0$
∴ Distance of the point (1, −2, 4) from the plane xy − 3 = 0

#### Question 2:

Show that the four points (0, −1, −1), (4, 5, 1), (3, 9, 4) and (−4, 4, 4) are coplanar and find the equation of the common plane.

The equation of the plane passing through the points (0, −1, −1), (4, 5, 1) and (3, 9, 4) is given by

#### Question 3:

Show that the following points are coplanar.
(i) (0, −1, 0), (2, 1, −1), (1, 1, 1) and (3, 3, 0)
(ii) (0, 4, 3), (−1, −5, −3), (−2, −2, 1) and (1, 1, −1)

(i) The equation of the plane passing through points (0, −1, 0), (2, 1, −1), (1, 1, 1) is given by

(ii) The equation of the plane passing through (0, 4, 3), (−1, −5, −3), (−2, −2, 1) is

#### Question 4:

Find the coordinates of the point where the line through (3,$-$4,$-$5) and  B (2,$-$3,1) crosses the plane passing through three points L(2,2,1), M(3,0,1) and N(4,$-$1,0). Also, find the ratio in which diveides the line segment AB.

Equation of the plane passing through the points L(2, 2, 1),  M(3, 0, 1) and N(4, −1, 0) is

The equation of line segment through A(3, −4, −5) and B(2−3, 1) is

Any point on this line is of the form .

This point lies on the plane (1).

$\therefore \left[\left(-\lambda +3\right)\stackrel{^}{i}+\left(\lambda -4\right)\stackrel{^}{j}+\left(6\lambda -5\right)\stackrel{^}{k}\right].\left(2\stackrel{^}{i}+\stackrel{^}{j}+\stackrel{^}{k}\right)=7\phantom{\rule{0ex}{0ex}}⇒2\left(-\lambda +3\right)+\left(\lambda -4\right)+\left(6\lambda -5\right)=7\phantom{\rule{0ex}{0ex}}⇒5\lambda =10\phantom{\rule{0ex}{0ex}}⇒\lambda =2$
Thus, the coordinates of the point P are (−2 + 3, 2 − 4, 6 × 2 − 5) i.e. (1, −2, 7).

Suppose P divides the line segment AB in the ratio μ : 1.

Thus, the point P divides the line segment AB externally in the ratio 2 : 1.

#### Question 1:

Find the distance between the parallel planes 2xy + 3z − 4 = 0 and 6x − 3y + 9z + 13 = 0.

#### Question 2:

Find the equation of the plane which passes through the point (3, 4, −1) and is parallel to the plane 2x − 3y + 5z + 7 = 0. Also, find the distance between the two planes.

#### Question 3:

Find the equation of the plane mid-parallel to the planes 2x − 2y + z + 3 = 0 and 2x − 2y + z + 9 = 0.

#### Question 4:

Find the distance between the planes

#### Question 1:

Find the angle between the line $\stackrel{\to }{r}=\left(2\stackrel{^}{i}+3\stackrel{^}{j}+9\stackrel{^}{k}\right)+\lambda \left(2\stackrel{^}{i}+3\stackrel{^}{j}+4\stackrel{^}{k}\right)$ and the plane $\stackrel{\to }{r}·\left(\stackrel{^}{i}+\stackrel{^}{j}+\stackrel{^}{k}\right)=5.$

#### Question 2:

Find the angle between the line $\frac{x-1}{1}=\frac{y-2}{-1}=\frac{z+1}{1}$ and the plane 2x + yz = 4.

#### Question 3:

Find the angle between the line joining the points (3, −4, −2) and (12, 2, 0) and the plane 3xy + z = 1.

#### Question 4:

The line $\stackrel{\to }{r}=\stackrel{^}{i}+\lambda \left(2\stackrel{^}{i}-m\stackrel{^}{j}-3\stackrel{^}{k}\right)$ is parallel to the plane $\stackrel{\to }{r}·\left(m\stackrel{^}{i}+3\stackrel{^}{j}+\stackrel{^}{k}\right)=4.$ Find m.

#### Question 5:

Show that the line whose vector equation is $\stackrel{\to }{r}=2\stackrel{^}{i}+5\stackrel{^}{j}+7\stackrel{^}{k}+\lambda \left(\stackrel{^}{i}+3\stackrel{^}{j}+4\stackrel{^}{k}\right)$ is parallel to the plane whose vector equation is $\stackrel{\to }{r}·\left(\stackrel{^}{i}+\stackrel{^}{j}-\stackrel{^}{k}\right)=7.$ Also, find the distance between them.

#### Question 6:

Find the vector equation of the line through the origin which is perpendicular to the plane $\stackrel{\to }{r}·\left(\stackrel{^}{i}+2\stackrel{^}{j}+3\stackrel{^}{k}\right)=3.$

#### Question 7:

Find the equation of the plane through (2, 3, −4) and (1, −1, 3) and parallel to x-axis.

#### Question 8:

Find the equation of a plane passing through the points (0, 0, 0) and (3, −1, 2) and parallel to the line $\frac{x-4}{1}=\frac{y+3}{-4}=\frac{z+1}{7}.$

#### Question 9:

Find the vector and Cartesian equations of the line passing through (1, 2, 3) and parallel to the planes

Disclaimer: The answer given for this problem in the text book is incorrect. The problem should be same as problem #19 to get the text book answer.

#### Question 10:

Prove that the line of section of the planes 5x + 2y − 4z + 2 = 0 and 2x + 8y + 2z − 1 = 0 is parallel to the plane 4x − 2y − 5z − 2 = 0.

#### Question 11:

Find the vector equation of the line passing through the point (1, −1, 2) and perpendicular to the plane 2xy + 3z − 5 = 0.

#### Question 12:

Find the equation of the plane through the points (2, 2, −1) and (3, 4, 2) and parallel to the line whose direction ratios are 7, 0, 6.

#### Question 13:

Find the angle between the line $\frac{x-2}{3}=\frac{y+1}{-1}=\frac{z-3}{2}$ and the plane 3x + 4y + z + 5 = 0.

#### Question 14:

Find the equation of the plane passing through the intersection of the planes x − 2y + z = 1 and 2x + y + z = 8 and parallel to the line with direction ratios proportional to 1, 2, 1. Also, find the perpendicular distance of (1, 1, 1) from this plane.

#### Question 15:

State when the line $\stackrel{\to }{r}=\stackrel{\to }{a}+\lambda \stackrel{\to }{b}$ is parallel to the plane $\stackrel{\to }{r}·\stackrel{\to }{n}=d.$ Show that the line $\stackrel{\to }{r}=\stackrel{^}{i}+\stackrel{^}{j}+\lambda \left(3\stackrel{^}{i}-\stackrel{^}{j}+2\stackrel{^}{k}\right)$ is parallel to the plane $\stackrel{\to }{r}·\left(2\stackrel{^}{j}+\stackrel{^}{k}\right)=3.$ Also, find the distance between the line and the plane.

#### Question 16:

Show that the plane whose vector equation is $\stackrel{\to }{r}·\left(\stackrel{^}{i}+2\stackrel{^}{j}-\stackrel{^}{k}\right)=1$ and the line whose vector equation is $\stackrel{\to }{r}=\left(-\stackrel{^}{i}+\stackrel{^}{j}+\stackrel{^}{k}\right)+\lambda \left(2\stackrel{^}{i}+\stackrel{^}{j}+4\stackrel{^}{k}\right)$ are parallel. Also, find the distance between them.

#### Question 17:

Find the equation of the plane through the intersection of the planes 3x − 4y + 5z = 10 and 2x + 2y − 3z = 4 and parallel to the line x = 2y = 3z.

#### Question 18:

Find the vector and Cartesian forms of the equation of the plane passing through the point (1, 2, −4) and parallel to the lines

$\stackrel{\to }{r}=\left(\stackrel{^}{i}+2\stackrel{^}{j}-4\stackrel{^}{k}\right)+\lambda \left(2\stackrel{^}{i}+3\stackrel{^}{j}+6\stackrel{^}{k}\right)$ and $\stackrel{\to }{r}=\left(\stackrel{^}{i}-3\stackrel{^}{j}+5\stackrel{^}{k}\right)+\mu \left(\stackrel{^}{i}+\stackrel{^}{j}-\stackrel{^}{k}\right)$. Also, find the distance of the point (9, −8, −10) from the plane thus obtained.      [CBSE 2014]

The equations of the given lines are

$\stackrel{\to }{r}=\left(\stackrel{^}{i}+2\stackrel{^}{j}-4\stackrel{^}{k}\right)+\lambda \left(2\stackrel{^}{i}+3\stackrel{^}{j}+6\stackrel{^}{k}\right)$

$\stackrel{\to }{r}=\left(\stackrel{^}{i}-3\stackrel{^}{j}+5\stackrel{^}{k}\right)+\mu \left(\stackrel{^}{i}+\stackrel{^}{j}-\stackrel{^}{k}\right)$

We know that the vector equation of a plane passing through a point $\stackrel{\to }{a}$ and parallel to $\stackrel{\to }{b}$ and $\stackrel{\to }{c}$ is given by $\left(\stackrel{\to }{r}-\stackrel{\to }{a}\right).\left(\stackrel{\to }{b}×\stackrel{\to }{c}\right)=0$.

Here, $\stackrel{\to }{a}=\stackrel{^}{i}+2\stackrel{^}{j}-4\stackrel{^}{k}$, $\stackrel{\to }{b}=2\stackrel{^}{i}+3\stackrel{^}{j}+6\stackrel{^}{k}$ and $\stackrel{\to }{c}=\stackrel{^}{i}+\stackrel{^}{j}-\stackrel{^}{k}$.

$\therefore \stackrel{\to }{b}×\stackrel{\to }{c}=\left|\begin{array}{ccc}\stackrel{^}{i}& \stackrel{^}{j}& \stackrel{^}{k}\\ 2& 3& 6\\ 1& 1& -1\end{array}\right|=-9\stackrel{^}{i}+8\stackrel{^}{j}-\stackrel{^}{k}$

So, the vector equation of the plane is

$\left(\stackrel{\to }{r}-\stackrel{\to }{a}\right).\left(\stackrel{\to }{b}×\stackrel{\to }{c}\right)=0\phantom{\rule{0ex}{0ex}}⇒\left[\stackrel{\to }{r}-\left(\stackrel{^}{i}+2\stackrel{^}{j}-4\stackrel{^}{k}\right)\right].\left(-9\stackrel{^}{i}+8\stackrel{^}{j}-\stackrel{^}{k}\right)=0\phantom{\rule{0ex}{0ex}}⇒\stackrel{\to }{r}.\left(-9\stackrel{^}{i}+8\stackrel{^}{j}-\stackrel{^}{k}\right)=\left(\stackrel{^}{i}+2\stackrel{^}{j}-4\stackrel{^}{k}\right).\left(-9\stackrel{^}{i}+8\stackrel{^}{j}-\stackrel{^}{k}\right)\phantom{\rule{0ex}{0ex}}⇒\stackrel{\to }{r}.\left(-9\stackrel{^}{i}+8\stackrel{^}{j}-\stackrel{^}{k}\right)=1×\left(-9\right)+2×8+\left(-4\right)×\left(-1\right)=-9+16+4=11$

Thus, the vector equation of the plane is $\stackrel{\to }{r}.\left(-9\stackrel{^}{i}+8\stackrel{^}{j}-\stackrel{^}{k}\right)=11$.

The Cartesian equation of this plane is

$\left(x\stackrel{^}{i}+y\stackrel{^}{j}+z\stackrel{^}{k}\right).\left(-9\stackrel{^}{i}+8\stackrel{^}{j}-\stackrel{^}{k}\right)=11\phantom{\rule{0ex}{0ex}}⇒-9x+8y-z=11$

Now,

Distance of the point (9, −8, −10) from the plane $-9x+8y-z=11$

= Length of perpendicular from (9, −8, −10) from the plane $-9x+8y-z-11=0$

#### Question 19:

Find the equation of the plane passing through the points (3, 4, 1) and (0, 1, 0) and parallel to the line $\frac{x+3}{2}=\frac{y-3}{7}=\frac{z-2}{5}.$

#### Question 20:

Find the coordinates of the point where the line $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{2}$ intersects the plane xy + z − 5 = 0. Also, find the angle between the line and the plane.

#### Question 21:

Find the vector equation of the line passing through (1, 2, 3) and perpendicular to the plane $\stackrel{\to }{r}·\left(\stackrel{^}{i}+2\stackrel{^}{j}-5\stackrel{^}{k}\right)+9=0.$

#### Question 22:

Find the angle between the line $\frac{x+1}{2}=\frac{y}{3}=\frac{z-3}{6}$ and the plane 10x + 2y − 11z = 3.

#### Question 23:

Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes

#### Question 24:

Find the value of λ such that the line $\frac{x-2}{6}=\frac{y-1}{\lambda }=\frac{z+5}{-4}$ is perpendicular to the plane 3xy − 2z = 7.

Disclaimer: It should be "parallel" instead of "perpendicular" in the given problem.

#### Question 25:

Find the equation of the plane passing through the points (−1, 2, 0), (2, 2, −1) and parallel to the line $\frac{x-1}{1}=\frac{2y+1}{2}=\frac{z+1}{-1}$.        [CBSE 2015]

The general equation of the plane passing through the point (−1, 2, 0) is given by

$a\left(x+1\right)+b\left(y-2\right)+c\left(z-0\right)=0$                 .....(1)

If this plane passes through the point (2, 2, −1), we have

Direction ratio's of the normal to the plane (1) are a, b, c.

The equation of the given line is $\frac{x-1}{1}=\frac{2y+1}{2}=\frac{z+1}{-1}$. This can be re-written as
$\frac{x-1}{1}=\frac{y+\frac{1}{2}}{1}=\frac{z+1}{-1}$

Direction ratio's of the line are 1, 1, −1.

The required plane is parallel to the given line when the normal to this plane is perpendicular to this line.

Solving (2) and (3), we get

$\frac{a}{0+1}=\frac{b}{-1+3}=\frac{c}{3-0}\phantom{\rule{0ex}{0ex}}⇒\frac{a}{1}=\frac{b}{2}=\frac{c}{3}=\lambda \left(\mathrm{Say}\right)\phantom{\rule{0ex}{0ex}}⇒a=\lambda ,b=2\lambda ,c=3\lambda$

Putting these values of a, b, c in (1), we have

$\lambda \left(x+1\right)+2\lambda \left(y-2\right)+3\lambda \left(z-0\right)=0\phantom{\rule{0ex}{0ex}}⇒x+1+2y-4+3z=0\phantom{\rule{0ex}{0ex}}⇒x+2y+3z=3$

Thus, the equation of the required plane is x + 2y + 3z = 3.

#### Question 26:

Find the equation of the plane that contains the point A (2, 1, $-$1) and is perpendicular to the line of intersection of the planes $2x+y-z=3$ and $x+2y+z=2$. Also, find the angle between the plane thus obtained and the y - axis.

Equation of plane that contains point
$⇒a\left(x-2\right)+b\left(y-1\right)+c\left(z+1\right)=0$
Since the plane is perpendicular to the intersection of the plane

Solving equations (1) and (2), we get
$a=-b=c=\lambda \left(\mathrm{say}\right)$
Thus, equation of a plane,
$\lambda \left(x-2\right)-\lambda \left(y-1\right)+\lambda \left(z+1\right)=0\phantom{\rule{0ex}{0ex}}x-y+z=0$
Let the angle between the plane and y-axis be $\theta$

Hence, the angle between the plane and y-axis is ${\mathrm{sin}}^{-1}\left(\frac{1}{\sqrt{3}}\right)$

#### Question 1:

Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the
(i) yz-plane
(ii) zx-plane

#### Question 2:

Find the coordinates of the point where the line through (3, −4, −5) and (2, −3, 1) crosses the plane 2x + y + z = 7.

#### Question 3:

Find the distance of the point (−1, −5, −10) from the point of intersection of the line $\stackrel{\to }{r}=\left(2\stackrel{^}{i}-\stackrel{^}{j}+2\stackrel{^}{k}\right)+\lambda \left(3\stackrel{^}{i}+4\stackrel{^}{j}+2\stackrel{^}{k}\right)$ and the plane $\stackrel{\to }{r}.\left(\stackrel{^}{i}-\stackrel{^}{j}+\stackrel{^}{k}\right)=5.$

#### Question 4:

Find the distance of the point (2, 12, 5) from the point of intersection of the line $\stackrel{\to }{r}=2\stackrel{^}{i}-4\stackrel{^}{j}+2\stackrel{^}{k}+\lambda \left(3\stackrel{^}{i}+4\stackrel{^}{j}+2\stackrel{^}{k}\right)$ and $\stackrel{\to }{r}.\left(\stackrel{^}{i}-2\stackrel{^}{j}+\stackrel{^}{k}\right)=0$.     [CBSE 2014]

The equation of the given line is $\stackrel{\to }{r}=2\stackrel{^}{i}-4\stackrel{^}{j}+2\stackrel{^}{k}+\lambda \left(3\stackrel{^}{i}+4\stackrel{^}{j}+2\stackrel{^}{k}\right)$.

The position vector of any point on the given line is

$\stackrel{\to }{r}=\left(2+3\lambda \right)\stackrel{^}{i}+\left(-4+4\lambda \right)\stackrel{^}{j}+\left(2+2\lambda \right)\stackrel{^}{k}$                      .....(1)

If this lies on the plane $\stackrel{\to }{r}.\left(\stackrel{^}{i}-2\stackrel{^}{j}+\stackrel{^}{k}\right)=0$, then

$\left[\left(2+3\lambda \right)\stackrel{^}{i}+\left(-4+4\lambda \right)\stackrel{^}{j}+\left(2+2\lambda \right)\stackrel{^}{k}\right].\left(\stackrel{^}{i}-2\stackrel{^}{j}+\stackrel{^}{k}\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(2+3\lambda \right)-2\left(-4+4\lambda \right)+\left(2+2\lambda \right)=0\phantom{\rule{0ex}{0ex}}⇒2+3\lambda +8-8\lambda +2+2\lambda =0\phantom{\rule{0ex}{0ex}}⇒3\lambda =12\phantom{\rule{0ex}{0ex}}⇒\lambda =4$

Putting $\lambda =4$ in (1), we get $\left(2+3×4\right)\stackrel{^}{i}+\left(-4+4×4\right)\stackrel{^}{j}+\left(2+2×4\right)\stackrel{^}{k}$ or $14\stackrel{^}{i}+12\stackrel{^}{j}+10\stackrel{^}{k}$ as the coordinate of the point of intersection of the given line and the plane.

The position vector of the given point is $2\stackrel{^}{i}+12\stackrel{^}{j}+5\stackrel{^}{k}$.

∴ Required distance = Distance between $14\stackrel{^}{i}+12\stackrel{^}{j}+10\stackrel{^}{k}$ and $2\stackrel{^}{i}+12\stackrel{^}{j}+5\stackrel{^}{k}$

#### Question 5:

Find the distance of the point P(−1, −5, −10) from the point of intersection of the line joining the points A(2, −1, 2) and B(5, 3, 4) with the plane $x-y+z=5$.                         [CBSE 2014, 2015]

The equation of the line passing through the points A(2, −1, 2) and B(5, 3, 4) is given by

The coordinates of any point on the line

$\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{2}=\lambda \left(\mathrm{say}\right)$ are $\left(3\lambda +2,4\lambda -1,2\lambda +2\right)$         .....(1)

If it lies on the plane $x-y+z=5$, then

$3\lambda +2-\left(4\lambda -1\right)+2\lambda +2=5\phantom{\rule{0ex}{0ex}}⇒\lambda +5=5\phantom{\rule{0ex}{0ex}}⇒\lambda =0$

Putting $\lambda =0$ in (1), we get (2, −1, 2) as the coordinates of the point of intersection of the given line and plane.

∴ Required distance = Distance between points (−1, −5, −10) and (2, −1, 2)

#### Question 6:

Find the distance of the point P(3, 4, 4) from the point, where the line joining the points A(3, −4, −5) and B(2, −3, 1) intersects the plane 2x + y + z = 7.                [CBSE 2015]

The equation of the line passing through the points A(3, −4, −5) and B(2, −3, 1) is given by

The coordinates of any point on the line

$\frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+5}{6}=\lambda \left(\mathrm{say}\right)$ are $\left(-\lambda +3,\lambda -4,6\lambda -5\right)$         .....(1)

If it lies on the plane 2x + y + z = 7, then

$2\left(-\lambda +3\right)+\left(\lambda -4\right)+\left(6\lambda -5\right)=7\phantom{\rule{0ex}{0ex}}⇒5\lambda -3=7\phantom{\rule{0ex}{0ex}}⇒5\lambda =10\phantom{\rule{0ex}{0ex}}⇒\lambda =2$

Putting $\lambda =2$ in (1), we get (1, −2, 7) as the coordinates of the point of intersection of the given line and plane.

∴ Required distance = Distance between points (3, 4, 4) and (1, −2, 7)

#### Question 7:

Find the distance of the point (1, $-$5, 9) from the plane $x-y+z=$5 measured along the line $x=y=z$.

The equation of line parallel to the line xyz and passing through the point (1, −5, 9) is

$\frac{x-1}{1}=\frac{y+5}{1}=\frac{z-9}{1}$     .....(1)

Any point on this line is of the form (k + 1, k − 5, k + 9).

If (k + 1, k − 5, k + 9) be the point of intersection of line (1) and the given plane, then

(k + 1) − (k − 5) + (k + 9) = 5

⇒ k = −10

So, the point of intersection of line (1) and the given plane is (−10 + 1, −10 − 5, −10 + 9) i.e. (−9, −15, −1).

∴ Required distance = Distance between (1, −5, 9) and (−9, −15, −1) = $\sqrt{{\left(1+9\right)}^{2}+{\left(-5+15\right)}^{2}+{\left(9+1\right)}^{2}}=\sqrt{3×{10}^{2}}=10\sqrt{3}$ units

#### Question 8:

Find the coordinates of the point where the line $\frac{x-1}{3}=\frac{y+4}{7}=\frac{z+4}{2}$ cuts the xy - plane.

Given: $\frac{x-1}{3}=\frac{y+4}{7}=\frac{z+4}{2}$
Let $\frac{x-1}{3}=\frac{y+4}{7}=\frac{z+4}{2}$ = K (say)

Since the given coordinates lie on $xy$-plane

Thus,
Hence, coordinates of the point are (7, 10, 0)

#### Question 9:

Find the distance of the point P() from the point of intersection Q of the line $\stackrel{\to }{r}=\left(3\stackrel{^}{i}-2\stackrel{^}{j}+6\stackrel{^}{k}\right)+\lambda \left(2\stackrel{^}{i}-\stackrel{^}{j}+2\stackrel{^}{k}\right)$ and the plane $\stackrel{\to }{r}=\left(\stackrel{^}{i}-\stackrel{^}{j}+\stackrel{^}{k}\right)=6.$ Also, write the vector equation of the line PQ.

Given: Line: $\stackrel{\to }{r}=\left(3+2\lambda \right)\stackrel{^}{i}-\left(2+\lambda \right)\stackrel{^}{j}+\left(6+2\lambda \right)\stackrel{^}{k}$
Plane: $\stackrel{\to }{r}·\left(\stackrel{^}{i}-\stackrel{^}{j}+\stackrel{^}{k}\right)=6$
Now, point of intersection
$\left[\left(3+2\lambda \right)\stackrel{^}{i}+\left(-2-\lambda \right)\stackrel{^}{j}+\left(6+2\lambda \right)\stackrel{^}{k}\right]·\left(\stackrel{^}{i}-\stackrel{^}{j}+\stackrel{^}{k}\right)=6\phantom{\rule{0ex}{0ex}}⇒3+2\lambda +2+\lambda +6+2\lambda =6\phantom{\rule{0ex}{0ex}}⇒\lambda =-1$
Thus point Q :
Point P
Hence, distance PQ = $\sqrt{{\left(1+2\right)}^{2}+{\left(-1+4\right)}^{2}+{\left(4-7\right)}^{2}}$
$=\sqrt{9+9+9}\phantom{\rule{0ex}{0ex}}=3\sqrt{3}$
Also, line equation in vector form PQ
PQ: $\stackrel{\to }{r}=-2\stackrel{^}{i}-4\stackrel{^}{j}+7\stackrel{^}{k}+K\left(3\stackrel{^}{i}+3\stackrel{^}{j}-3\stackrel{^}{k}\right)$

#### Question 1:

Write the equation of the plane whose intercepts on the coordinate axes are 2, −3 and 4.

#### Question 2:

Reduce the equations of the following planes to intercept form and find the intercepts on the coordinate axes.
(i) 4x + 3y − 6z − 12 = 0
(ii) 2x + 3y − z = 6
(iii) 2xy + z = 5

#### Question 3:

Find the equation of a plane which meets the axes at A, B and C, given that the centroid of the triangle ABC is the point (α, β, γ).

#### Question 4:

Find the equation of the plane passing through the point (2, 4, 6) and making equal intercepts on the coordinate axes.

#### Question 5:

A plane meets the coordinate axes at A, B and C, respectively, such that the centroid of triangle ABC is (1, −2, 3). Find the equation of the plane.

#### Question 1:

Show that the lines are coplanar. Also, find the equation of the plane containing them.

#### Question 2:

Show that the lines are coplanar. Also, find the equation of the plane containing them.

#### Question 3:

Find the equation of the plane containing the line $\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}$ and the point (0, 7, −7) and show that the line $\frac{x}{1}=\frac{y-7}{-3}=\frac{z+7}{2}$ also lies in the same plane.

#### Question 4:

Find the equation of the plane which contains two parallel lines

#### Question 5:

Show that the lines $\frac{x+4}{3}=\frac{y+6}{5}=\frac{z-1}{-2}$ and 3x − 2y + z + 5 = 0 = 2x + 3y + 4z − 4 intersect. Find the equation of the plane in which they lie and also their point of intersection.

#### Question 6:

Show that the plane whose vector equation is $\stackrel{\to }{r}·\left(\stackrel{^}{i}+2\stackrel{^}{j}-\stackrel{^}{k}\right)=3$ contains the line whose vector equation is $\stackrel{\to }{r}=\stackrel{^}{i}+\stackrel{^}{j}+\lambda \left(2\stackrel{^}{i}+\stackrel{^}{j}+4\stackrel{^}{k}\right).$

#### Question 7:

Find the equation of the plane determined by the intersection of the lines

#### Question 8:

Find the vector equation of the plane passing through the points (3, 4, 2) and (7, 0, 6) and perpendicular to the plane 2x − 5y − 15 = 0. Also, show that the plane thus obtained contains the line $\stackrel{\to }{r}=\stackrel{^}{i}+3\stackrel{^}{j}-2\stackrel{^}{k}+\lambda \left(\stackrel{^}{i}-\stackrel{^}{j}+\stackrel{^}{k}\right).$

#### Question 9:

If the lines are perpendicular, find the value of k and, hence, find the equation of the plane containing these lines.

#### Question 10:

Find the coordinates of the point where the line $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{2}$ intersect the plane xy + z − 5 = 0. Also, find the angle between the line and the plane.

#### Question 11:

Find the vector equation of the plane passing through three points with position vectors Also, find the coordinates of the point of intersection of this plane and the line $\stackrel{\to }{r}=3\stackrel{^}{i}-\stackrel{^}{j}-\stackrel{^}{k}+\lambda \left(2\stackrel{^}{i}-2\stackrel{^}{j}+\stackrel{^}{k}\right).$

#### Question 12:

Show that the lines $\frac{5-x}{-4}=\frac{y-7}{4}=\frac{z+3}{-5}$ and $\frac{x-8}{7}=\frac{2y-8}{2}=\frac{z-5}{3}$ are coplanar.         [CBSE 2014]

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

$\frac{x-5}{4}=\frac{y-7}{4}=\frac{z+3}{-5}$

and $\frac{x-8}{7}=\frac{y-4}{1}=\frac{z-5}{3}$

We know that the lines $\frac{x-{x}_{1}}{{l}_{1}}=\frac{y-{y}_{1}}{{m}_{1}}=\frac{z-{z}_{1}}{{n}_{1}}$ and $\frac{x-{x}_{2}}{{l}_{2}}=\frac{y-{y}_{2}}{{m}_{2}}=\frac{z-{z}_{2}}{{n}_{2}}$ are coplanar if $\left|\begin{array}{ccc}{x}_{2}-{x}_{1}& {y}_{2}-{y}_{1}& {z}_{2}-{z}_{1}\\ {l}_{1}& {m}_{1}& {n}_{1}\\ {l}_{2}& {m}_{2}& {n}_{2}\end{array}\right|=0$.

Here,

${x}_{1}=5,{y}_{1}=7,{z}_{1}=-3,{x}_{2}=8,{y}_{2}=4,{z}_{2}=5\phantom{\rule{0ex}{0ex}}{l}_{1}=4,{m}_{1}=4,{n}_{1}=-5,{l}_{2}=7,{m}_{2}=1,{n}_{2}=3$

$\therefore \left|\begin{array}{ccc}{x}_{2}-{x}_{1}& {y}_{2}-{y}_{1}& {z}_{2}-{z}_{1}\\ {l}_{1}& {m}_{1}& {n}_{1}\\ {l}_{2}& {m}_{2}& {n}_{2}\end{array}\right|\phantom{\rule{0ex}{0ex}}=\left|\begin{array}{ccc}8-5& 4-7& 5-\left(-3\right)\\ 4& 4& -5\\ 7& 1& 3\end{array}\right|\phantom{\rule{0ex}{0ex}}=\left|\begin{array}{ccc}3& -3& 8\\ 4& 4& -5\\ 7& 1& 3\end{array}\right|\phantom{\rule{0ex}{0ex}}=3\left(12+5\right)+3\left(12+35\right)+8\left(4-28\right)\phantom{\rule{0ex}{0ex}}=51+141-192\phantom{\rule{0ex}{0ex}}=0$

So, the given lines are coplanar.

#### Question 13:

Find the equation of a plane which passes through the point (3, 2, 0) and contains the line $\frac{x-3}{1}=\frac{y-6}{5}=\frac{z-4}{4}$.         [CBSE 2015]

Let the equation of the plane passing through (3, 2, 0) be

$a\left(x-3\right)+b\left(y-2\right)+c\left(z-0\right)=0$                     .....(1)

The line $\frac{x-3}{1}=\frac{y-6}{5}=\frac{z-4}{4}$ passes through the point (3, 6, 4) and its direction ratios are proportional to 1, 5, 4.

If plane (1) contains this line, then it must pass through (3, 6, 4) and must be parallel to the line.

Also,

Solving (2) and (3), we get

$\frac{a}{4-5}=\frac{b}{1-0}=\frac{c}{0-1}\phantom{\rule{0ex}{0ex}}⇒\frac{a}{-1}=\frac{b}{1}=\frac{c}{-1}=\lambda \left(\mathrm{Say}\right)\phantom{\rule{0ex}{0ex}}⇒a=-\lambda ,b=\lambda ,c=-\lambda$

Putting these values of a, b, c in (1), we get

$-\lambda \left(x-3\right)+\lambda \left(y-2\right)-\lambda \left(z-0\right)=0\phantom{\rule{0ex}{0ex}}⇒-x+3+y-2-z=0\phantom{\rule{0ex}{0ex}}⇒-x+y-z+1=0\phantom{\rule{0ex}{0ex}}⇒x-y+z-1=0$

Thus, the equation of the required plane is xy + z − 1 = 0.

#### Question 14:

Show that the lines $\frac{x+3}{-3}=\frac{y-1}{1}=\frac{z-5}{5}$ and $\frac{x+1}{-1}=\frac{y-2}{2}=\frac{z-5}{5}$ are coplanar. Hence, find the equation of the plane containing these lines.

The lines $\frac{x-{x}_{1}}{{a}_{1}}=\frac{y-{y}_{1}}{{b}_{1}}=\frac{z-{z}_{1}}{{c}_{1}}$ and $\frac{x-{x}_{2}}{{a}_{2}}=\frac{y-{y}_{2}}{{b}_{2}}=\frac{z-{z}_{2}}{{c}_{2}}$ are coplanar if $\left|\begin{array}{ccc}{x}_{2}-{x}_{1}& {y}_{2}-{y}_{1}& {z}_{2}-{z}_{1}\\ {a}_{1}& {b}_{1}& {c}_{1}\\ {a}_{2}& {b}_{2}& {c}_{2}\end{array}\right|=0$.

The given lines are $\frac{x+3}{-3}=\frac{y-1}{1}=\frac{z-5}{5}$ and $\frac{x+1}{-1}=\frac{y-2}{2}=\frac{z-5}{5}$.

Now, $\left|\begin{array}{ccc}{x}_{2}-{x}_{1}& {y}_{2}-{y}_{1}& {z}_{2}-{z}_{1}\\ {a}_{1}& {b}_{1}& {c}_{1}\\ {a}_{2}& {b}_{2}& {c}_{2}\end{array}\right|=$$\left|\begin{array}{ccc}-1-\left(-3\right)& 2-1& 5-5\\ -3& 1& 5\\ -1& 2& 5\end{array}\right|=\left|\begin{array}{ccc}2& 1& 0\\ -3& 1& 5\\ -1& 2& 5\end{array}\right|=2\left(5-10\right)-1\left(-15+5\right)+0=-10+10+0=0$

So, the given lines are coplanar.

The equation of the plane containing the given lines is

$\left|\begin{array}{ccc}x-\left(-3\right)& y-1& z-5\\ -3& 1& 5\\ -1& 2& 5\end{array}\right|=0\phantom{\rule{0ex}{0ex}}⇒\left|\begin{array}{ccc}x+3& y-1& z-5\\ -3& 1& 5\\ -1& 2& 5\end{array}\right|=0\phantom{\rule{0ex}{0ex}}⇒\left(x+3\right)\left(5-10\right)-\left(y-1\right)\left(-15+5\right)+\left(z-5\right)\left(-6+1\right)=0\phantom{\rule{0ex}{0ex}}⇒-5\left(x+3\right)+10\left(y-1\right)-5\left(z-5\right)=0\phantom{\rule{0ex}{0ex}}⇒x-2y+z=0$
Thus, the equation of the plane containing the given lines is x − 2yz = 0.

#### Question 15:

If the line$\frac{x-3}{2}=\frac{y+2}{-1}=\frac{z+4}{3}$ lies in the plane $lx+my-z=$9, then find the value of ${l}^{2}+{m}^{2}$.

The line $\frac{x-{x}_{1}}{a}=\frac{y-{y}_{1}}{b}=\frac{z-{z}_{1}}{c}$ lies in the plane AxByCz D = 0 iff (i) $A{x}_{1}+B{y}_{1}+C{z}_{1}+D=0$ and (ii) $aA+bB+cC=0$.

It is given that the line $\frac{x-3}{2}=\frac{y+2}{-1}=\frac{z+4}{3}$ lies in the plane $lx+my-z=9$.

Also,

Solving (1) and (2), we get

l = 1 and m = −1

∴ l2m2 = 12 + (−1)2 = 1 + 1 = 2

Thus, the value of l2 + m2 is 2.

#### Question 16:

Find the values of $\lambda$ for which the lines $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z+3}{{\lambda }^{2}}$ and $\frac{x-3}{1}=\frac{y-2}{{\lambda }^{2}}=\frac{z-1}{2}$ are coplanar.

The lines $\frac{x-{x}_{1}}{{a}_{1}}=\frac{y-{y}_{1}}{{b}_{1}}=\frac{z-{z}_{1}}{{c}_{1}}$ and $\frac{x-{x}_{2}}{{a}_{2}}=\frac{y-{y}_{2}}{{b}_{2}}=\frac{z-{z}_{2}}{{c}_{2}}$ are coplanar if $\left|\begin{array}{ccc}{x}_{2}-{x}_{1}& {y}_{2}-{y}_{1}& {z}_{2}-{z}_{1}\\ {a}_{1}& {b}_{1}& {c}_{1}\\ {a}_{2}& {b}_{2}& {c}_{2}\end{array}\right|=0$.

The given lines $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z+3}{{\lambda }^{2}}$ and $\frac{x-3}{1}=\frac{y-2}{{\lambda }^{2}}=\frac{z-1}{2}$ are coplanar.

$\therefore \left|\begin{array}{ccc}{x}_{2}-{x}_{1}& {y}_{2}-{y}_{1}& {z}_{2}-{z}_{1}\\ {a}_{1}& {b}_{1}& {c}_{1}\\ {a}_{2}& {b}_{2}& {c}_{2}\end{array}\right|=0$
$⇒\left|\begin{array}{ccc}3-1& 2-2& 1-\left(-3\right)\\ 1& 2& {\lambda }^{2}\\ 1& {\lambda }^{2}& 2\end{array}\right|=0\phantom{\rule{0ex}{0ex}}⇒\left|\begin{array}{ccc}2& 0& 4\\ 1& 2& {\lambda }^{2}\\ 1& {\lambda }^{2}& 2\end{array}\right|=0\phantom{\rule{0ex}{0ex}}⇒2\left(4-{\lambda }^{4}\right)-0+4\left({\lambda }^{2}-2\right)=0\phantom{\rule{0ex}{0ex}}⇒-2{\lambda }^{4}+4{\lambda }^{2}=0$

Thus, the values of $\lambda$ are 0, $-\sqrt{2}$ and $\sqrt{2}$.

#### Question 17:

If the lines $x=$5,  $\frac{y}{3-\alpha }=\frac{z}{-2}$ and $x=\alpha$$\frac{y}{-1}=\frac{z}{2-\alpha }$ are coplanar, find the values of $\alpha$.

The lines $\frac{x-{x}_{1}}{{a}_{1}}=\frac{y-{y}_{1}}{{b}_{1}}=\frac{z-{z}_{1}}{{c}_{1}}$ and $\frac{x-{x}_{2}}{{a}_{2}}=\frac{y-{y}_{2}}{{b}_{2}}=\frac{z-{z}_{2}}{{c}_{2}}$ are coplanar if $\left|\begin{array}{ccc}{x}_{2}-{x}_{1}& {y}_{2}-{y}_{1}& {z}_{2}-{z}_{1}\\ {a}_{1}& {b}_{1}& {c}_{1}\\ {a}_{2}& {b}_{2}& {c}_{2}\end{array}\right|=0$.

The given lines $\frac{x-5}{0}=\frac{y}{3-\alpha }=\frac{z}{-2}$ and $\frac{x-\alpha }{0}=\frac{y}{-1}=\frac{z}{2-\alpha }$ are coplanar.

$\therefore \left|\begin{array}{ccc}\alpha -5& 0-0& 0-0\\ 0& 3-\alpha & -2\\ 0& -1& 2-\alpha \end{array}\right|=0\phantom{\rule{0ex}{0ex}}⇒\left|\begin{array}{ccc}\alpha -5& 0& 0\\ 0& 3-\alpha & -2\\ 0& -1& 2-\alpha \end{array}\right|=0\phantom{\rule{0ex}{0ex}}⇒\left(\alpha -5\right)\left[\left(3-\alpha \right)×\left(2-\alpha \right)-2\right]-0+0=0\phantom{\rule{0ex}{0ex}}⇒\left(\alpha -5\right)\left({\alpha }^{2}-5\alpha +4\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(\alpha -5\right)\left(\alpha -1\right)\left(\alpha -4\right)=0$

Thus, the values of $\alpha$ are 1, 4 and 5.

#### Question 18:

If the straight lines $\frac{x-1}{2}=\frac{y+1}{k}=\frac{z}{2}$ and $\frac{x+1}{2}=\frac{y+1}{2}=\frac{z}{k}$ are coplanar, find the equations of the planes containing them.

The lines $\frac{x-{x}_{1}}{{a}_{1}}=\frac{y-{y}_{1}}{{b}_{1}}=\frac{z-{z}_{1}}{{c}_{1}}$ and $\frac{x-{x}_{2}}{{a}_{2}}=\frac{y-{y}_{2}}{{b}_{2}}=\frac{z-{z}_{2}}{{c}_{2}}$ are coplanar if $\left|\begin{array}{ccc}{x}_{2}-{x}_{1}& {y}_{2}-{y}_{1}& {z}_{2}-{z}_{1}\\ {a}_{1}& {b}_{1}& {c}_{1}\\ {a}_{2}& {b}_{2}& {c}_{2}\end{array}\right|=0$.

The given lines  $\frac{x-1}{2}=\frac{y+1}{k}=\frac{z}{2}$ and $\frac{x+1}{2}=\frac{y+1}{2}=\frac{z}{k}$ are coplanar.

$\therefore \left|\begin{array}{ccc}-1-1& -1-\left(-1\right)& 0-0\\ 2& k& 2\\ 2& 2& k\end{array}\right|=0\phantom{\rule{0ex}{0ex}}⇒\left|\begin{array}{ccc}-2& 0& 0\\ 2& k& 2\\ 2& 2& k\end{array}\right|=0\phantom{\rule{0ex}{0ex}}⇒-2\left({k}^{2}-4\right)-0+0=0\phantom{\rule{0ex}{0ex}}⇒{k}^{2}-4=0\phantom{\rule{0ex}{0ex}}⇒k=±2$
The equation of the plane containing the given lines is $\left|\begin{array}{ccc}x-1& y+1& z\\ 2& k& 2\\ 2& 2& k\end{array}\right|=0$.

For k = 2,  $\left|\begin{array}{ccc}x-1& y+1& z\\ 2& k& 2\\ 2& 2& k\end{array}\right|=$$\left|\begin{array}{ccc}x-1& y+1& z\\ 2& 2& 2\\ 2& 2& 2\end{array}\right|=0$
So, no plane exists for k = 2.

For k = −2,
$\left|\begin{array}{ccc}x-1& y+1& z\\ 2& k& 2\\ 2& 2& k\end{array}\right|=0$
$⇒\left|\begin{array}{ccc}x-1& y+1& z\\ 2& -2& 2\\ 2& 2& -2\end{array}\right|=0\phantom{\rule{0ex}{0ex}}⇒\left(x-1\right)\left(4-4\right)-\left(y+1\right)\left(-4-4\right)+z\left(4+4\right)=0\phantom{\rule{0ex}{0ex}}⇒8\left(y+1\right)+8z=0\phantom{\rule{0ex}{0ex}}⇒y+z+1=0$
Thus, the equation of the plane containing the given lines is yz + 1 = 0.

#### Question 19:

Find the vector equation of the plane that contains the lines  and the point (-1, 3, -4). Also, find the length of the perpendicular drawn from the point (2, 1, 4) to the plane, thus obtained.

Given line $\stackrel{\to }{r}=\left(\stackrel{^}{i}+\stackrel{^}{j}\right)+\lambda \left(\stackrel{^}{i}+2\stackrel{^}{j}-\stackrel{^}{k}\right)$ passes through (1,1,0) and is parallel to the vector $\stackrel{^}{i}+2\stackrel{^}{j}-\stackrel{^}{k}$

i.e. The given plane passes through (1,1,0) and B(−1, 3, −4) and is parallel to $\stackrel{\to }{b}=\stackrel{\to }{c}+2\stackrel{^}{j}-\stackrel{^}{k}$

Let $\stackrel{\to }{n}$ be the normal vector to the required plane.

Then $\stackrel{\to }{n}$ is perpendicular to both

i.e. $\stackrel{\to }{n}$ is parallel to $\stackrel{\to }{AB}×\stackrel{\to }{B}$

Then the required plane passes through $\stackrel{\to }{\alpha }=\stackrel{^}{i}+\stackrel{^}{j}$ and is perpendicular to ${\stackrel{\to }{n}}_{1}=6\stackrel{^}{i}-6\stackrel{^}{j}-6\stackrel{^}{k}$
So, its vector equation is

Equation of plane is x – y – z = 0
The length of perpendicular form (2, 1, 4) to plane x – y – z = 0 is

#### Question 1:

Find the shortest distance between the lines

#### Question 2:

Find the shortest distance between the lines

#### Question 3:

Find the shortest distance between the lines $\frac{x-1}{2}=\frac{y-3}{4}=\frac{z+2}{1}$ and $3x-y-2z+4=0=2x+y+z+1$.

The equation of the plane containing the line $3x-y-2z+4=0=2x+y+z+1$ is

If it is parallel to the line $\frac{x-1}{2}=\frac{y-3}{4}=\frac{z+2}{1}$, then

$2\left(3+2\lambda \right)+4\left(\lambda -1\right)+\left(\lambda -2\right)=0\phantom{\rule{0ex}{0ex}}⇒9\lambda =0\phantom{\rule{0ex}{0ex}}⇒\lambda =0$

Putting $\lambda =0$ in (1), we get

This is the equation of the plane containing the second line and parallel to the first line.

Now, the line $\frac{x-1}{2}=\frac{y-3}{4}=\frac{z+2}{1}$ passes through (1, 3, −2).

∴ Shortest distance between the given lines

= Length of the perpendicular from (1, 3, −2) to the plane $3x-y-2z+4=0$

#### Question 1:

Find the image of the point (0, 0, 0) in the plane 3x + 4y − 6z + 1 = 0.

#### Question 2:

Find the reflection of the point (1, 2, −1) in the plane 3x − 5y + 4z = 5.

#### Question 3:

Find the coordinates of the foot of the perpendicular drawn from the point (5, 4, 2) to the line $\frac{x+1}{2}=\frac{y-3}{3}=\frac{z-1}{-1}.$ Hence, or otherwise, deduce the length of the perpendicular.

#### Question 4:

Find the image of the point with position vector $3\stackrel{^}{i}+\stackrel{^}{j}+2\stackrel{^}{k}$ in the plane $\stackrel{\to }{r}·\left(2\stackrel{^}{i}-\stackrel{^}{j}+\stackrel{^}{k}\right)=4.$ Also, find the position vectors of the foot of the perpendicular and the equation of the perpendicular line through $3\stackrel{^}{i}+\stackrel{^}{j}+2\stackrel{^}{k}.$

#### Question 5:

Find the coordinates of the foot of the perpendicular from the point (1, 1, 2) to the plane 2x − 2y + 4z + 5 = 0. Also, find the length of the perpendicular.

#### Question 6:

Find the distance of the point (1, −2, 3) from the plane xy + z = 5 measured along a line parallel to $\frac{x}{2}=\frac{y}{3}=\frac{z}{-6}.$

The given plane is x − y + z = 5. We need to find the distance of the point from the point(1, -2, 3) measured along a parallel line $\frac{x}{2}=\frac{y}{3}=\frac{z}{-6}.$
Let the line from point be P(1, -2, 3) and meet the plane at point Q.
Direction ratios of the line from the point ​(1, -2, 3) to the given plane will be the same as the given line $\frac{x}{2}=\frac{y}{3}=\frac{z}{-6}.$
So the equation of the line passing through P and with same direction ratios will be:

coordinates of any point on the line PQ are .
Now, since Q lies on the plane so it must satisfy the equation of the plane.
that is, ​x − y + z = 5

therefore, 2λ+1 - (3λ - 2) + (-6λ +3) = 5

coordinates of Q are  =
using distance formula we have the length of PQ as

Hence PQ = 1

So, ​the distance of the point (1, −2, 3) from the plane x − y + z = 5  is 1

#### Question 7:

Find the coordinates of the foot of the perpendicular from the point (2, 3, 7) to the plane 3xyz = 7. Also, find the length of the perpendicular.

#### Question 8:

Find the image of the point (1, 3, 4) in the plane 2xy + z + 3 = 0.

#### Question 9:

Find the distance of the point with position vector $-\stackrel{^}{i}-5\stackrel{^}{j}-10\stackrel{^}{k}$ from the point of intersection of the line $\stackrel{\to }{r}=\left(2\stackrel{^}{i}-\stackrel{^}{j}+2\stackrel{^}{k}\right)+\lambda \left(3\stackrel{^}{i}+4\stackrel{^}{j}+12\stackrel{^}{k}\right)$ with the plane $\stackrel{\to }{r}·\left(\stackrel{^}{i}-\stackrel{^}{j}+\stackrel{^}{k}\right)=5.$

#### Question 10:

Find the length and the foot of the perpendicular from the point (1, 1, 2) to the plane $\stackrel{\to }{r}·\left(\stackrel{^}{i}-2\stackrel{^}{j}+4\stackrel{^}{k}\right)+5=0.$

Disclaimer: The answer given for this problem in the text book is incorrect.

#### Question 11:

Find the coordinates of the foot of the perpendicular and the perpendicular distance of the  point P (3, 2, 1) from the plane 2xy + z + 1 = 0. Also, find the image of the point in the plane.

#### Question 12:

Find the direction cosines of the unit vector perpendicular to the plane $\stackrel{\to }{r}·\left(6\stackrel{^}{i}-3\stackrel{^}{j}-2\stackrel{^}{k}\right)+1=0$ passing through the origin.

For the unit vector perpendicular to the given plane, we need to convert the given equation of plane into normal form.

#### Question 13:

Find the coordinates of the foot of the perpendicular drawn from the origin to the plane 2x − 3y + 4z − 6 = 0.

#### Question 14:

Find the length and the foot of perpendicular from the point $\left(1,\frac{3}{2},2\right)$ to the plane $2x-2y+4z+5=0$.          [NCERT EXEMPLAR]

Let M be the foot of the perpendicular from P$\left(1,\frac{3}{2},2\right)$ on the plane $2x-2y+4z+5=0$.

Then, PM is the normal to the plane. So, its direction ratios are proportional to 2, −2, 4.

Since PM passes through P$\left(1,\frac{3}{2},2\right)$, therefore, its equation is

$\frac{x-1}{2}=\frac{y-\frac{3}{2}}{-2}=\frac{z-2}{4}=\lambda \left(\mathrm{Say}\right)$

Let the coordinates of M be $\left(2\lambda +1,-2\lambda +\frac{3}{2},4\lambda +2\right)$.

Now, M lies on the plane $2x-2y+4z+5=0$.

$\therefore 2\left(2\lambda +1\right)-2\left(-2\lambda +\frac{3}{2}\right)+4\left(4\lambda +2\right)+5=0\phantom{\rule{0ex}{0ex}}⇒24\lambda +12=0\phantom{\rule{0ex}{0ex}}⇒\lambda =-\frac{1}{2}$

So, the coordinates of M are $\left(2×\left(-\frac{1}{2}\right)+1,-2×\left(-\frac{1}{2}\right)+\frac{3}{2},4×\left(-\frac{1}{2}\right)+2\right)$ or $\left(0,\frac{5}{2},0\right)$.

Thus, the coordinates of the foot of the perpendicular are $\left(0,\frac{5}{2},0\right)$.

Now,

$\mathrm{PM}=\sqrt{{\left(1-0\right)}^{2}+{\left(\frac{3}{2}-\frac{5}{2}\right)}^{2}+{\left(2-0\right)}^{2}}=\sqrt{1+1+4}=\sqrt{6}$

Thus, the length of the perpendicular from the given point to the plane is $\sqrt{6}$ units.

#### Question 15:

Find the position vector of the foot of perpendicular and the perpendicular distance from the point P with position vector $2\stackrel{^}{i}+3\stackrel{^}{j}+4\stackrel{^}{k}$ to the plane $\stackrel{\to }{r}.\left(2\stackrel{^}{i}+\stackrel{^}{j}+3\stackrel{^}{k}\right)-26=0$. Also find image of P in the plane.