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

Let Q be the image of the point P ( 0 , 0 , 0 ) in the plane 3 x + 4 y - 6 z + 1 = 0 . Then P Q , is normal to the plane. So, the direction ratios of P Q are proportional to 3 , 4 , - 6 . Since P Q passes through P ( 0 , 0 , 0 ) and has direction ratios proportional to 3 , 4 and - 6 , equation of P Q is x - 0 3 = y - 0 4 = z - 0 - 6 = r ( say) Let the Co-ordinate of Q be ( 3 r , 4 r , - 6 r ) . Let R be the mid-point R = 0 + 3 r 2 , 0 + 4 r 2 , 0 - 6 r 2 = 3 r 2 , 2 r , - 3 r Since R lies in the plane 3 x + 4 y - 6 z + 1 = 0 , 3 3 r 2 + 4 2 r - 6 - 3 r + 1 = 0 ⇒ r = - 2 61 Substituting this in the coordinates of Q , we get Q = - 6 61 , - 8 61 , 12 61 .

Find the distance of the point with position vector - i ^ - 5 j ^ - 10 k ^ from the point of intersection of the line r → = 2 i ^ - j ^ + 2 k ^ + λ 3 i ^ + 4 j ^ + 12 k ^ with the plane r → · i ^ - j ^ + k ^ = 5 .

The equation of the line is given as r → = 2 i ^ - j ^ + 2 k ^ + λ 3 i ^ + 4 j ^ + 2 k ^ and the plane is given as r → · i ^ - j ^ + k ^ = 5 . Now substitute the line equation in the plane equation, 2 i ^ - j ^ + 2 k ^ + λ 3 i ^ + 4 j ^ + 2 k ^ · i ^ - j ^ + k ^ = 5 2 i ^ - j ^ + 2 k ^ . i ^ - j ^ + k ^ + λ 3 i ^ + 4 j ^ + 2 k ^ . i ^ - j ^ + k ^ = 5 i → . i → = j → . j → = k → . k → = 1 , i → . j → = j → . i → = k → . j → = k → . i → = j → . k → = i → . k → = 0 2 + 1 + 2 + λ 3 - 4 + 2 = 5 5 + λ = 5 λ = 0 So, the equation of the line can be reduced to r → = 2 i ^ - j ^ + 2 k ^ . If r → = x i ^ + y j ^ + z k ^ , then the coordinates can hence be obtained as 2 , - 1 , 2 . So, the vector joining the points - 1 , - 5 , - 10 and 2 , - 1 , 2 can be written as P → = 2 - - 1 i ^ + - 1 - 5 j ^ + 2 - - 10 k ^ P → = 3 i ^ + 4 j ^ + 12 k ^ The distance can be found using the magnitude of the vector as P → = 3 2 + 4 2 + 12 2 = 9 + 16 + 144 = 169 = 13

Find the length and the foot of perpendicular from the point 1 , 3 2 , 2 to the plane 2 x - 2 y + 4 z + 5 = 0 .

Let M be the foot of the perpendicular from P 1 , 3 2 , 2 on the plane 2 x - 2 y + 4 z + 5 = 0 . Then, P M is normal to the plane. So, its direction ratios are proportional to coefficients of x , y , and z i.e., 2 , - 2 , 4 Since P M passes through P 1 , 3 2 , 2 , therefore, its equation is x - 1 2 = y - 3 2 - 2 = z - 2 4 = λ say ⇒ x = 2 λ + 1 , y = - 2 λ + 3 2 , and z = 4 λ + 2 Let the coordinates of M be 2 λ + 1 , - 2 λ + 3 2 , 4 λ + 2 Now, M lies on the plane 2 x - 2 y + 4 z + 5 = 0 ∴ 2 ( 2 λ + 1 ) - 2 - 2 λ + 3 2 + 4 ( 4 λ + 2 ) + 5 = 0 ⇒ 24 λ + 12 = 0 ⇒ λ = - 1 2 So, the coordinates of M are 2 × - 1 2 + 1 , - 2 × - 1 2 + 3 2 , 4 × - 1 2 + 2 i.e., 0 , 5 2 , 0 , which is the required foot of the perpendicular from the point P 1 , 3 2 , 2 to the plane 2 x - 2 y + 4 z + 5 = 0 . Now, the length of the perpendicular from the point P 1 , 3 2 , 2 to the point M 0 , 5 2 , 0 is given by P M = 1 - 0 2 + 3 2 - 5 2 2 + 2 2 = 1 + 1 + 4 = 6

E M B I B E

MATHEMATICS CLASS XII VOLUME-2>Chapter 10 - The Plane>EXERCISE

R D Sharma Solutions for Chapter: The Plane, Exercise 15: EXERCISE

Author:R D Sharma

R D Sharma Mathematics Solutions for Exercise - R D Sharma Solutions for Chapter: The Plane, Exercise 15: EXERCISE

Attempt the free practice questions on Chapter 10: The Plane, Exercise 15: EXERCISE with hints and solutions to strengthen your understanding. MATHEMATICS CLASS XII VOLUME-2 solutions are prepared by Experienced Embibe Experts.

Questions from R D Sharma Solutions for Chapter: The Plane, Exercise 15: EXERCISE with Hints & Solutions

MEDIUM

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-2>Chapter 10 - The Plane>EXERCISE>Q 1

Find the image of the point $R = (\frac{0 + 3 r}{2}, \frac{0 + 4 r}{2}, \frac{0 - 6 r}{2}) = (\frac{3 r}{2}, 2 r, - 3 r)$ in the plane $3 x + 4 y - 6 z + 1 = 0 .$

EASY

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-2>Chapter 10 - The Plane>EXERCISE>Q 2

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

EASY

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-2>Chapter 10 - The Plane>EXERCISE>Q 4

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

EASY

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-2>Chapter 10 - The Plane>EXERCISE>Q 5

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

EASY

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-2>Chapter 10 - The Plane>EXERCISE>Q 8

Find the image of the point $(1, 3, 4)$ in the plane $2 x - y + z + 3 = 0 .$

EASY

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-2>Chapter 10 - The Plane>EXERCISE>Q 9

Find the distance of the point with position vector $- \hat{i} - 5 \hat{j} - 10 \hat{k}$ from the point of intersection of the line $\vec{r} = (2 \hat{i} - \hat{j} + 2 \hat{k}) + λ (3 \hat{i} + 4 \hat{j} + 12 \hat{k})$ with the plane $\vec{r} \cdot (\hat{i} - \hat{j} + \hat{k}) = 5 .$

MEDIUM

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-2>Chapter 10 - The Plane>EXERCISE>Q 14

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

MEDIUM

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-2>Chapter 10 - The Plane>EXERCISE>Q 15

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

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