Question

Let $L$ be the line of intersection of the planes $2x+3y+z=1$ and $x+3y+2z=2$. If $L$ makes an angle $\alpha$ with the positive $X$-axis, then $\cos \alpha$ equals

• A. $\frac{1}{\sqrt{3}}$
• B. $\frac{1}{\sqrt{2}}$
• C. $\frac{1}{\sqrt{5}}$
• D. 1

Answer 1

The correct option is A $\frac{1}{\sqrt{3}}$

We have $2x+3y+z=1$
$\Rightarrow 2x+3y=1-z\cdots \left(i\right)$
$\Rightarrow x+3y+2z=2$
$\Rightarrow x+3y=2-2z\cdots \left(ii\right)$
$\left(i\right)-\left(ii\right)$

$\Rightarrow 2x+3y-x-3y=1-z-2+2z$
$\Rightarrow x=-1+z$
$\Rightarrow z=\frac{x+1}{1}\cdots \left(iii\right)$

putting $x$ in $\left(ii\right)$
$\Rightarrow -1+z+3y=2-2z$
$\Rightarrow 3z=3-3y$
$\Rightarrow z=\frac{y-1}{-1}\cdots \left(iv\right)$
From $\left(iii\right)\left(iv\right)$
$\frac{x+1}{1}=\frac{y-1}{-1}=\frac{z}{1}$
Thus angle between above line and $X$ axis having DR's $(1,0,0)$ will be
$\cos \alpha =\frac{1\cdot 1+0+0}{\sqrt{1+1+1}\sqrt{1}}=\frac{1}{\sqrt{3}}$