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please solve RD sharma class 12 chapter 28 The Plane exercise 28.8 question 5 maths textbook solution

Answers (1)

The answer of the given question is 28x-17y+9z=0.

Hint:- We know that, equation of a plane passing through the line of intersection of two planes

a_{1}x+b_{1}y+c_{1}z+d_{1}=0 and a_{2}x+b_{2}y+c_{2}z+d_{2}=0  is given by

\left (a_{1}x+b_{1}y+c_{1}z+d_{1} \right )+k\left (a_{2}x+b_{2}y+c_{2}z+d_{2} \right )=0

 

Given:- 2x-y=0 and 3z-y=0  and plane 4x+5y-3z=8

Solution:-  So equation of plane passing through the line of intersection of given two planes is

2x-y=0 and 3z-y=0  is
\left (2x-y \right )+k \left (3z-y \right )=0 \\ 2x-y+ 3kz-ky=0 \\ x(2)+y(-1-k)+z(3k)=0                                                          …(i)
We know that, two planes are perpendicular if

a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}=0                                                                       …(ii)

Given, plane (i) is perpendicular to plane

4x+5y-3z=8                                             …(iii)
Using (i) and (iii) in equation (ii)

(2)(4)+(-1-k)(5)+(3k)(-3)=0\\ 3-14k=0\\ k=\frac{3}{14}

Putting value of k in equation (i)

\begin{gathered} x(2)+y(-1-k)+z(3 k)=0 \\ \end{gathered}

\begin{gathered} x(2)+y\left(-1-\frac{3}{14}\right)+z\left(3\left(\frac{3}{14}\right)\right)=0 \\ \end{gathered}

\begin{gathered} x(2)+y\left(-\frac{17}{14}\right)+z\left(\frac{9}{14}\right)=0 \end{gathered}

Multiplying with 14 we get

28x-17y+9z=0

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