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Q.
If the angle $\theta$ between the line $\frac{x+1}{1}=\frac{y-1}{2}=\frac{z-2}{2}$ and the plane $2x-y+\sqrt{p} z+4=0$ is such that $sin\,\theta =\frac{1}{3}$ , then the value of $p$ is
Given equation of line is $\frac{x+1}{1}=\frac{y-1}{2}=\frac{z-2}{2}$
Direction cosines of the line are,
$<\,l,m, n >\,= <\, 1,2,2>\,$
and equation of plane is $2 x-y+\sqrt{p} z+4=0$
Direction cosines of the plane are,
$<\, a, b, c >\,=<\, 2,-1, \sqrt{p} >$
Also given, $\sin \,\theta=\frac{1}{3}$
$\therefore \, \sin \theta=\left|\frac{a l+b m+c n}{\sqrt{a^{2}+b^{2}+c^{2}} \sqrt{1^{2}+m^{2}+n^{2}}}\right|$
$\Rightarrow \, \frac{1}{3}=| \frac{(2)(1)+(-1)(2)+(\sqrt{p})(2)}{\sqrt{4+1+p} \sqrt{1+4+4} |}$
$\Rightarrow \, \frac{1}{3}=\left|\frac{2-2+2 \sqrt{p}}{\sqrt{p+5} \sqrt{9}}\right|=\left|\frac{2 \sqrt{p}}{3 \sqrt{p+5}}\right|$
On squaring both sides, we get
$\frac{1}{9}=\frac{4}{9} \times \frac{p}{(p+5)}$
$\Rightarrow \,p+5=4 p$
$ \Rightarrow \,3 p=5$
$ \Rightarrow \,p=\frac{5}{3}$