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  • please solve RD sharma class 12 chapter 28 The Plane exercise 28.8 question 3 maths textbook solution

please solve RD sharma class 12 chapter 28 The Plane exercise 28.8 question 3 maths textbook solution

Answers (1)

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 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 )+\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
(2x-7y+4z-3)+k(3x-5y+4z+11)=0\\ (2x-7y+4z-3)+3kx-5ky+4kz+11k=0\\ x(2+3k)+y(-7-5k)+z(4+4k)-3+11k=0\\                          … (i)

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

(-2)(2+3k)+(1)(-7-5k)+(3)(4+4k)-3+11k=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 \\ \end{gathered}

\begin{gathered} 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 \\ \end{gathered}

\begin{gathered} 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

15x-47y+28z-7=0

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