3D Geometry 3 Question 15
####15. If an angle between the line, $\frac{x+1}{2}=\frac{y-2}{1}=\frac{z-3}{-2}$ and the plane,
(2019 Main, 12 Jan II) $x-2 y-k z=3$ is $\cos ^{-1} \frac{2 \sqrt{2}}{3}$, then value of $k$ is
(a) $\sqrt{\frac{5}{3}}$
(b) $\sqrt{\frac{3}{5}}$
(c) $-\frac{3}{5}$
(d) $-\frac{5}{3}$
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Answer:
(a)
Solution:
- Clearly, direction ratios of given line are
$2,1,-2$ and direction ratios of normal to the given plane are $1,-2,-k$.
As we know angle ’ $\theta$ ’ between line and plane can be obtained by
$ \begin{gathered} \sin \theta=\frac{\left|a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}\right|}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}} \\ \text { So, } \sin \theta=\frac{|2-2+2 k|}{\sqrt{4+1+4} \sqrt{1+4+k^{2}}} \\ \Rightarrow \sin \cos ^{-1} \frac{2 \sqrt{2}}{3}=\frac{|2 k|}{3 \sqrt{5+k^{2}}} \quad \text { given } \theta=\cos ^{-1} \frac{2 \sqrt{2}}{3} \\ \Rightarrow \frac{2|k|}{3 \sqrt{5+k^{2}}}=\frac{1}{3} \because \cos \theta=\frac{2 \sqrt{2}}{3} \Rightarrow \sin \theta=\frac{1}{3} \Rightarrow \theta=\sin ^{-1} \frac{1}{3} \\ \Rightarrow \cos ^{-1} \frac{2 \sqrt{2}}{3}=\sin ^{-1} \frac{1}{3} \\ \Rightarrow \quad \Delta^{-3} \\ \quad 4 k^{2}=5+k^{2} \Rightarrow 3 k^{2}=5 \\ k= \pm \sqrt{\frac{5}{3}} \end{gathered} $
