Question

Let $P_1$ be a plane containing two lines $L_1: \frac{x-2}{3}=\frac{y}{0}=\frac{z-4}{-1}$ and $L_2: \frac{x-7}{-2}=\frac{y}{0}=\frac{z-2}{1}$ $P_2$ be another plane containing a triangle whose vertices are $(3,-2,0),(2,0,0)$ and $(0,5,0)$. $P$ is the point of intersection of $L _1=0$ and $L _2=0$.
If acute angle between the lines $L _1=0$ and $P _1=0= P _2$ is $\cot ^{-1}(\lambda)$ then $\lambda$ is

• A. 3
• B. 5
• C. 7
• D. 10

Answer 1

Answer: (A)

$ L _1: \frac{ x -2}{3}=\frac{ y }{0}=\frac{ z -4}{-1}=l$
$\Rightarrow(3 l+2,0,-l+4) $
$L _2: \frac{ x -7}{-2}=\frac{ y }{0}=\frac{ z -2}{1}= r $
$\Rightarrow(-2 r +7,0,+ r +2)$
For point of intersection of $L _1=0 \& L _2=0$
$3l + 2 = -2r + 7 $
$-l + 4 = r + 2$
$3l + 2r = 5$ …(i)
$ l + r = 2$ …(ii)
From (i) & (ii)
$l = r = 1$
$\therefore P \equiv(5,0,3)$
Equation of plane $P _1$ containing $L _1=0 \& L _2=0$ is -
$\begin{vmatrix}x-2 & y & z-4 \\ 3 & 0 & -1 \\ -2 & 0 & 1\end{vmatrix}=0 \Rightarrow y=0$
Equation of plane $P _2$ containing points $(3,-2,0),(2,0,0) \&(0,5,0)$ is $z =0$ i.e. $x$-y plane.
Line of intersection of $x$ - $y$ plane \& $z$ - $x$ plane is $x$-axis i.e. $\frac{ x }{ a }=\frac{ y }{0}=\frac{ z }{0}$
Acute angle between $L_1=0$ and $P_1=0=P_2$ ( $x$-axis)
$\cos \theta=\frac{3 \times 1+0 \times 0+(-1) \times 0}{\sqrt{9+0+1} \sqrt{1+0+0}}$
Dr's of $L_1=0: 3,0,-1$
$Dr ^{\prime}$ s of $x$-axis : $1,0,0$
$\cos \theta=\frac{3}{\sqrt{10}} \Rightarrow \cot \theta=3$
$ \Rightarrow \theta=\cot ^{-1} 3=\cot ^{-1} \lambda $
$\therefore \lambda=3 $