The line of shortest distance between the lines $\frac{x-2}{0}=\frac{y-1}{1}=\frac{z}{1}$ and $\frac{x-3}{2}=\frac{y-5}{2}=\frac{z-1}{1}$ makes an angle of $\cos ^{-1}\left(\sqrt{\frac{2}{27}}\right)$ with the plane $\mathrm{P}: \mathrm{a} x-y-z=0$, $(a>0)$. If the image of the point $(1,1,-5)$ in the plane $P$ is $(\alpha, \beta, \gamma)$, then $\alpha+\beta-\gamma$ is equal to _________________.
Answer (integer)
3
Solution
DR's of line of shortest distance<br/><br/>
$$
\left|\begin{array}{lll}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
0 & 1 & 1 \\
2 & 2 & 1
\end{array}\right|=-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}
$$<br/><br/>
angle between line and plane is $\cos ^{-1} \sqrt{\frac{2}{27}}=\alpha$<br/><br/>
$\cos \alpha=\sqrt{\frac{2}{27}}, \sin \alpha=\frac{5}{3 \sqrt{3}}$<br/><br/>
DR's normal to plane $(1,-1,-1)$<br/><br/>
$$
\sin \alpha=\left|\frac{-a-2+2}{\sqrt{4+4+1} \sqrt{a^2+1+1}}\right|=\frac{5}{3 \sqrt{3}}
$$<br/><br/>
$\sqrt{3}|a|=5 \sqrt{a^2+2}$<br/><br/>
$3 a^2=25 a^2+50$<br/><br/>
No value of (a)
About this question
Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios
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