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  • Need solution for RD Sharma maths class 12 chapter The Plane exercise 28.8 question 7

Need solution for RD Sharma maths class 12 chapter The Plane exercise 28.8 question 7

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Answer:-  The answer of the given question is x-10y-5z=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:-x+2y+3z+4=0  and  x-y+z+3=0

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

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

                \begin{array}{r} (x+2 y+3 z+4)+k(x-y+z+3)=0 \\\\ x(1+k)+y(2-k)+z(3+k)+4+3 k=0 \end{array} .          … (i)

Equation (i) is passing through origin, so       

\begin{gathered} (1+k)+(0)(2-k)+(0)(3+k)+4+3 k=0 \\\\ 0+0+0+4+3 k=0 \\\\ k=-\frac{4}{3} \end{gathered}

 

Put the value of k in equation (i)

 

        \begin{gathered} x(1+k)+y(2-k)+z(3+k)+4+3 k=0 \\\\ x\left(1-\frac{4}{3}\right)+y\left(2+\frac{4}{3}\right)+z\left(3-\frac{4}{3}\right)+4-\frac{12}{3}=0 \end{gathered}

        \begin{gathered} x\left(\frac{3-4}{3}\right)+y\left(\frac{6+4}{3}\right)+z\left(\frac{9-4}{3}\right)+4-4=0 \\\\ -\frac{x}{3}+\frac{10 y}{3}+\frac{5 z}{3}=0 \end{gathered}

Multiplying by 3, we get

        -x+10y+5z=0

        x-10y-5z=0

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