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Please solve RD Sharma class 12 chapter 28 The Plane exercise multiple choice question 5 maths textbook solution

Answers (1)

Answer:

 Option (b)

Hint:

 Solve the equation simultaneously.

Given:

 (1, 3, 4) in the plane 2x - y + z + 3 = 0

Solution:

We know, if the image of a point P(x0, y0, z0) on a plane Ax + By + Cz + D = 0 . . . (1) is Q(x1, y1, z1) then

 

The given plane is 2x - y + z + 3 = 0  . . . . .(2)

By comparing equations (2) and (3)

A = 2, B = -1, C = 1, D = 3

And here x_0 = 1, y_0 = 3, z_0 = 4

So

\begin{aligned} &\frac{x_1-1}{2}=\frac{y_1-3}{-1}=\frac{z_1-4}{1}=\frac{-2[(2\times 1)+((-1)\times 3)+(1\times 4)]}{2^2+(-1)^2+1^2}\\ &\frac{x_1-1}{2}=\frac{y_1-3}{-1}=\frac{z_1-4}{1}=\frac{-2(2-3+4+3)}{4+1+1}\\ &\frac{x_1-1}{2}=\frac{y_1-3}{-1}=\frac{z_1-4}{1}=\frac{-2\times 6}{6}\\ &\frac{x_1-1}{2}=\frac{y_1-3}{-1}=\frac{z_1-4}{1}=-2 \end{aligned}

Therefor,

x1 = -4 + 1

    =-3

y1 = 2 + 3

    =5

z1 = -2 + 4

    =2

So, the image of the (1, 3, 4), in the plane 2x - y + 3 = 0 , is (-3, 5, 2)

Posted by

Gurleen Kaur

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