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

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

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Answer:-  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 \text { 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

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

The plane passes through the point (2, 2,1). Therefore, this point will satisfy equation (1)
            \begin{gathered} \therefore(3 \times 2-2+2 \times 1-4)+\alpha(2+2+1-2)=0 \\\\ 2+3 \alpha=0 \\ \alpha=-\frac{2}{3} \end{gathered}

Substituting \alpha=-\frac{2}{3}  in equation (i), we obtain

            \begin{gathered} (3 x-y+2 z-4)-\frac{2}{3}(x+y+z-2)=0 \\\\ 3(3 x-y+2 z-4)-2(x+y+z-2)=0 \\\\ (9 x-3 y+6 z-12)-2(x+y+z-2)=0 \\\\ 7 x-5 y+4 z-8=0 \end{gathered}

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