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  • Need solution for RD Sharma maths class 12 chapter The Plane exercise 28.8 question 11

Need solution for RD Sharma maths class 12 chapter The Plane exercise 28.8 question 11

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Answer:-  The answer of the given question is 7x+13y+4z=9.

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:-2x+3y-z+1=0 and  x+y-2z+3=0

            Plane :3x-y-2z-4=0

Solution:-  Cartesian form of  equation of plane through the line of intersection of plane is

                A_{1} x+B_{1} y+C_{1} z+D_{1}+\lambda\left(A_{2} x+B_{2} y+C_{2} z+D_{2}\right)=0                       … (i)

Here the standard equation of plane is A_{1} x+B_{1} y+C_{1} z+D_{1} \text { and } A_{2} x+B_{2} y+C_{2} z+D_{2}

Substituting the values in equation (i) we get

            2 x+3 y-z+1+\lambda(x+y-2 z+3)=0

            (2+\lambda) x+(3+\lambda) y+(-1-2 \lambda) z+1+3 \lambda=0                    ...........(ii)

It is given that the plane 3x-y-2z-4=0 is perpendicular to the plane

We know that \emptyset=90^{\circ} where \cos 90^{\circ}=0 so we get

            A_{1} A_{2}+B_{1} B_{2}+C_{1} C_{2}=0                                                          … (iii)

By comparing the standard equation in Cartesian form\begin{array}{lll} A_{1}=2+\lambda, & B_{1}=3+\lambda, & C_{1}=-1-2 \lambda \\\\ A_{2}=3, & B_{2}=-1, & C_{2}=-2 \end{array}

Substituting these values in equation (iii)

            (2+\lambda) \cdot 3+(3+\lambda)(-1)+(-1-2 \lambda)(-2)=0

On further calculation

            \begin{gathered} 6+3 \lambda-3-\lambda+2+4 \lambda=0 \\ \lambda=-\frac{5}{6} \end{gathered}

Substituting the value of λ in equation (ii)

            \left(2+\frac{-5}{6}\right) x+\left(3+\frac{-5}{6}\right) y+\left(-1-2 \cdot \frac{-5}{6}\right) z+1+3 \cdot \frac{-5}{6}=0

By taking LCM

            \left(\frac{12-5}{6}\right) x+\left(\frac{18-5}{6}\right) y+\left(\frac{-6+10}{6}\right) z+\frac{6-15}{6}=0

We get

            7x+13y+4z-9=0

            7x+13y+4z=9

Posted by

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