Question

Let a plane $P_1$ containing the line $L_1:\frac{x-1}{1}=\frac{y-1}{-1}=\frac{z-1}{-1}$ and another plane $P_2$ containing the line $L_1$ and passses through the point $(0,1,1).$ If d.r's of normal to the plane $P_1$ are $1,1,0,$ then the acute angle between the planes $P_1$ and $P_2$ is equal to

• A. $\frac{\pi }{3}$
• B. $\frac{\pi }{6}$
• C. $\frac{\pi }{4}$
• D. $\frac{\pi }{8}$

Answer 1

The correct option is A $\frac{\pi }{3}$

Given line, $L_1:\frac{x-1}{1}=\frac{y-1}{-1}=\frac{z-1}{-1}$
Since plane $P_1$ containing the line $L_1$ and dr's normal to the plane $P_1$ is $(1,1,0)$
$\therefore$ Equation of plane $P_1$ is
$1(x-1)+1(y-1)+0(z-1)=0$
$\Rightarrow x-y-2=0$
Equation of plane $P_2$ containing the line $L_1$ and passing through $(0,1,1)$ is
$\left|\begin{array}{ccc}x-1& y-1& z-1\\ 1-0& 1-1& 1-1\\ 1& -1& -1\end{array}\right|=0$
$\Rightarrow \left|\begin{array}{ccc}x-1& y-1& z-1\\ 1& 0& 0\\ 1& -1& -1\end{array}\right|=0$
$\Rightarrow (y-1)(-1-0)+(z-1)(-1-0)=0$
$\Rightarrow y-z=0$
Let $\theta$ be the acute angle between $P_1$ and $P_2$.
Then, $\cos \theta =\frac{|1\cdot 0+1\cdot 1+0\cdot (-1)|}{\sqrt{2}\cdot \sqrt{2}}=\frac{1}{2}$
$\therefore \theta =\frac{\pi }{3}$