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)

$L_1:\frac{x-2}{3}=\frac{y}{0}=\frac{z-4}{-1}=l$
$\Rightarrow (3l+2,0,-l+4)$
$L_2:\frac{x-7}{-2}=\frac{y}{0}=\frac{z-2}{1}=r$
$\Rightarrow (-2r+7,0,+r+2)$
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 -
$\left|\begin{array}{ccc}x-2& y& z-4\\ 3& 0& -1\\ -2& 0& 1\end{array}\right|=0\Rightarrow y=0$
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 $L_1=0:3,0,-1$
$D{r}^{\prime }$ s of $x$-axis : $1,0,0$
$\cos \theta =\frac{3}{\sqrt{10}}\Rightarrow \cot \theta =3$
$\Rightarrow \theta ={\cot }^{-1}3={\cot }^{-1}\lambda$
$\therefore \lambda =3$