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Provide Solution for RD Sharma Class 12 Chapter 28 The Plane Exercise 28.11 Question 17

Answers (1)

Answer: $x-20 y+27 z-14=0$

Hint: Use Properties of plane

Given: $3 x-4 y+5 z=10 \text { and } 2 x+2 y-3 z=4 \text { and line } x=2 y=3 z$

Solution: We know that equation of plane passing through the intersection of plane $a_{1} x+b_{1} y+c_{1} z+d_{1}=0 \text { and } a_{2} x+b_{2} y+c_{2} z+d_{2}=0$ is given by

$\left(a_{1} x+b_{1} y+c_{1} z+d_{1}\right)+k\left(a_{2} x+b_{2} y+c_{2} z+d_{2}\right)=0$

So, equation of plane passing through the intersection of planes

$\begin{aligned} &3 x-4 y+5 z=10 \text { and } 2 x+2 y-3 z=4 \text { is } \\ \end{aligned}$

$(3 x-4 y+5 z-10)+k(2 x+2 y-3 z-4)=0 \\$ ………………. (1)

$\Rightarrow x(3+2 k)+y(-4+2 k)+z(5-3 k)+(-10-4 k)=0$

We know that line

$\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}$ is parallel to plane $a_{2} x+b_{2} y+c_{2} z+d_{2}=0 \text { if } a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0$

Given the plane is parallel to line $x=2 y=3 z$

Or, $\frac{x}{6}=\frac{y}{3}=\frac{z}{2}$ (dividing the equation by 6)

$\begin{aligned} &6 \times(3+2 k)+3 \times(-4+2 k)+2 \times(5-3 k)=0 \\ & \end{aligned}$

$\Rightarrow 18+12 k-12+6 k+10-6 k=0 \\$

$\Rightarrow k=\frac{-16}{12}=\frac{-4}{3}$

Putting the value of k in equation (1) we get

$\begin{aligned} &(3 x-4 y+5 z-10)-\frac{4}{3}(2 x+2 y-3 z-4)=0 \\ & \end{aligned}$

$\Rightarrow 3 x-\frac{8}{3} x-4 y-\frac{8}{3} y+5 z+4 z-10+\frac{16}{3}=0$

$\begin{aligned} &\Rightarrow \frac{x}{3}-\frac{20 y}{3}+9 z-\frac{14}{3}=0 \\ & \end{aligned}$

$\Rightarrow x-20 y+27 z-14=0$

The required equation is $x-20 y+27 z-14=0$

Posted by

infoexpert27

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