Question

The angle between the line $\frac{x+1}{2}=\frac{y}{3}=\frac{z-3}{6}$ and the plane $10x+2y-11z=3$ is ${\cos }^{-1}\left(\frac{p}{3q}\right)-\frac{\pi }{2},$ then which of the following statement(s) is/are correct?
(where $p,q$ are relative prime, $[.]$ denotes $G.I.F.$)

• A. $p-q=1$
• B. $2p+q=1$
• C. $\left[\sqrt{pq}\right]=7$
• D. $\left[\frac{pq}{p+q}\right]=3$

Answer 1

Answer: (A), (C), (D)

$\left[\frac{pq}{p+q}\right]=3$

Given :
Line: $\frac{x+1}{2}=\frac{y}{3}=\frac{z-3}{6},$
$D.r^{\prime }s$ of line $(a_1,b_1,c_1)=(2,3,6)$
Plane: $10x+$ $2y-11z=3$
$D.r^{\prime }s$ of normal to the plane $(a_2,b_2,c_2)=(10,2,-11)$
If the angle between the line and plane is $\theta$
$\Rightarrow \sin \theta =\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}$
$\cos (90-\theta )=$
$\frac{2\times 10+3\times 2+6\times (-11)}{\sqrt{(2^2+3^2+6^2)({10}^2+2^2+{11}^2)}}=\frac{-8}{21}$
$\therefore \theta ={90}^{\circ }-{\cos }^{-1}\left(\frac{-8}{21}\right)$
$\text{or}\theta =\frac{\pi }{2}-(\pi -{\cos }^{-1}\left(\frac{8}{21}\right))={\cos }^{-1}\left(\frac{8}{21}\right)-\frac{\pi }{2}\because {\cos }^{-1}(-x)=\pi -{\cos }^{-1}x$
$\Rightarrow p=8,q=7$