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  • Provide solution for RD sharma maths class 12 chapter 28 The Plane exercise 28.8 question 7

Provide solution for RD sharma maths class 12 chapter 28 The Plane exercise 28.8 question 7

Answers (1)

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 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\\ \left (x+2y+3z+4 \right )+k\left (x-y+z+3 \right )=0\\ x(1+k)+y(2-k)+z(3+k)+4+3k=0                                    … (i)

Equation (i) is passing through origin, so

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

Put the value of k in equation (i)

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

\begin{gathered} 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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