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  • Need solution for RD sharma maths class 12 chapter 28 The Plane exercise 28.8 question 15

Need solution for RD sharma maths class 12 chapter 28 The Plane exercise 28.8 question 15

Answers (1)

The answer of the given question is  7x-5y+4z-8=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:- 3x-y+2z=4 and x+y+z=2 , Point (2,2,1)

Solution:- The equation of any plane through the intersection of the planes is

3x-y+2z-4=0  and  x+y+z-2=0 is

\left (3x-y+2z-4 \right )+\alpha \left (x+y+z-2 \right )=0 , where \alpha \in \mathbb{R}                   …(i)

The plane passes through the point (2, 2,1). Therefore, this point will satisfy equation (1)

\therefore \left (3 \times 2-2+2 \times 1-4 \right )+\alpha \left (2+2+1-2 \right )=0\\ 2+3\alpha =0\\ \alpha =-\frac{2}{3}

Substituting \alpha =-\frac{2}{3} in equation (i), we obtain
\left (3x-y+2z-4 \right )-\frac{3}{2}\left (x+y+z-2 \right )=0\\ 3\left (3x-y+2z-4 \right )-2\left (x+y+z-2 \right )=0\\ \left (9x-3y+6z-12 \right )-2\left (x+y+z-2 \right )=0\\ 7x-5y+4z-8=0

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