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  • Need solution for RD Sharma maths class 12 chapter The Plane exercise 28.8 question 3

Need solution for RD Sharma maths class 12 chapter The Plane exercise 28.8 question 3

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Answer:-  The answer of the given question is 15x-47y+28z-7=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:-2x-7y+4z-3=0 and 3x-5y+4z+11=0

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

        \begin{aligned} &(2 x-7 y+4 z-3)+k(3 x-5 y+4 z+11)=0 \\\\ &2 x-7 y+4 z-3+3 k x-5 k y+4 k z+11 k=0 \\\\ &x(2+3 k)+y(-7-5 k)+z(4+4 k)-3+11 k=0 \end{aligned}                      … (i)

As given that, plane (i) is passing through the point (-2,1,3) so it satisfy the equation (i)

        (-2)(2+3 k)+(1)(-7-5 k)+(3)(4+4 k)-3+11 k=0

                                                                                                          k=\frac{1}{6}

Put the value of k in equation (i)

        \begin{gathered} x(2+3 k)+y(-7-5 k)+z(4+4 k)-3+11 k=0 \\\\ x\left(2+\frac{3}{6}\right)+y\left(-7-\frac{5}{6}\right)+z\left(4+\frac{4}{6}\right)-3+\frac{11}{6}=0 \end{gathered}

        \begin{gathered} x\left(\frac{12+3}{6}\right)+y\left(\frac{-42-5}{6}\right)+z\left(\frac{24+4}{6}\right)-\frac{18+11}{6}=0 \\\\ x\left(\frac{15}{6}\right)+y\left(-\frac{47}{6}\right)+z\left(\frac{28}{6}\right)-\frac{7}{6}=0 \end{gathered}

Multiplying by 6 we get

              15 x-47 y+28 z-7=0

Posted by

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