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Need solution for RD Sharma Maths Class 12 Chapter 28 The Plane Excercise 28.11 Question 20

Answers (1)

Answer: Coordinates of intersection is $(2,-1,2), \sin ^{-1}\left(\frac{1}{\sqrt{27}}\right)$

Hint: Use formula $\sin \theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}$

Given: $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{2} \text { and } x-y+z-5=0$

Solution: Let $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{2}=r$

$\Rightarrow x=3 r+2, y=4 r-1, z=2 r+2$

Substituting in the equation of the plane $x-y+z-5=0$ , We get,

$\begin{aligned} &(3 r+2)-(4 r-1)+(2 r+2)-5=0 \\ & \end{aligned}$

$\Rightarrow 3 r+2-4 r+1+2 r+2-5=0 \\$

$\Rightarrow r=0$

$\begin{aligned} &x=3 \times 0+2 \Rightarrow x=2 \\ \end{aligned}$

$y=4 \times 0-1 \Rightarrow y=-1 \\$

$z=2 \times 0+2 \Rightarrow z=2$

Hence, the coordinates of intersection is $(2,-1,2)$

Direction ratios of the line are $3, 4, 2$

Direction ratios of the line perpendicular to the plane are $1, -1, 1$

$\begin{aligned} &\sin \theta=\frac{3 \times 1+(-1) \times 4+2 \times 1}{\sqrt{3^{2}+4^{2}+2^{2}} \sqrt{1^{2}+(-1)^{2}+1^{2}}} \\ & \end{aligned}$

$\sin \theta=\frac{3-4+2}{\sqrt{9+16+4} \sqrt{1+1+1}}$

$\begin{aligned} &=\frac{1}{\sqrt{29} \sqrt{3}} \\ \end{aligned}$

$=\frac{1}{\sqrt{87}}$

$\theta=\sin ^{-1}\left(\frac{1}{\sqrt{87}}\right)$

The angle between the plane and line is $\sin ^{-1}\left(\frac{1}{\sqrt{87}}\right)$

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