Question

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

• A. $0$
• B. $\frac{1}{3}$
• C. $\frac{2}{3}$
• D. $\frac{4}{3}$
• E. $\frac{5}{3}$

Answer 1

Answer: (E)

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 $2x-y+\sqrt{p}z+4=0$
Direction cosines of the plane are,
$<a,b,c>=<2,-1,\sqrt{p}>$
Also given, $\sin \theta =\frac{1}{3}$
$\therefore \sin \theta =\left|\frac{al+bm+cn}{\sqrt{a^2+b^2+c^2}\sqrt{1^2+m^2+n^2}}\right|$
$\Rightarrow \frac{1}{3}=|\frac{(2)(1)+(-1)(2)+(\sqrt{p})(2)}{\sqrt{4+1+p}\sqrt{1+4+4}|}$
$\Rightarrow \frac{1}{3}=\left|\frac{2-2+2\sqrt{p}}{\sqrt{p+5}\sqrt{9}}\right|=\left|\frac{2\sqrt{p}}{3\sqrt{p+5}}\right|$
On squaring both sides, we get
$\frac{1}{9}=\frac{4}{9}\times \frac{p}{(p+5)}$
$\Rightarrow p+5=4p$
$\Rightarrow 3p=5$
$\Rightarrow p=\frac{5}{3}$