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  • Explain solution RD sharma class 12 chapter 28 The Plane exercise 28.8, question 18

Explain solution RD sharma class 12 chapter 28 The Plane exercise 28.8, question 18

Answers (1)

The answer of the given question are

  1. 13x+14y+11z=0, This plane doesn’t satisfies the given condition.
  2. The equation of the required plane is, 7x+11y+14z=15.
  3. The equation of the required plane is, 7x+11y+4z=33

Hint:-  By using intercept form of plane

            \frac{x}{a}+\frac{y}{b}+\frac{c}{z}=1

Given:-  The equation of the family of planes passing through the intersection of the planes

x+2y+3z-4=0 and 2x+y-z+5=0

Solution:- The equation of the family of planes passing through the intersection of the planes

        x+2y+3z-4=0 and 2x+y-z+5=0

               \left (x+2y+3z-4 \right )+k\left (2x+y-z+5 \right )=0, where k is some constant.

               (2k+1)x+(k+2)y+(3-k)z=4-5k\\ \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 )}

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 )\\

(4-5k)(3-k)=(4k+2)(4-5k)

(4-5k)(3-k-4k-2)=0

(4-5k)(1-5k)=0            

Either
             4-5k=0 \;\;\;or \;\;\;\;\; 1-5k=0\\ k=\frac{4}{5} \;\;\;\;\; or \;\;\;\;\; k=\frac{1}{5}

When k=\frac{4}{5}, the equation of the plane is

               \left (2 \times \frac{4}{5}+1 \right )x+\left (\frac{4}{5}+2 \right )y+\left ( 3-\frac{4}{5} \right )z=4-5\times \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 \times \frac{1}{5}+1 \right )x+\left (\frac{1}{5}+2 \right )y+\left ( 3-\frac{1}{5} \right )z=4-5\times \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(x-2)+11(y-3)+14(z+1)=0\\ 7x+11y+4z=33

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