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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

Answers (1)

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

infoexpert26

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