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. $\frac{1}{2}$
• C. $1$
• D. $\frac{1}{\sqrt{2}}$

Answer 1

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

We have

$2x+3y+z=1$

$⟹2x+3y=1-z$ ...(i)

$x+3y+2z=2$

$⟹x+3y=2-2z$ ...(ii)

$\left(i\right)-\left(ii\right)$

$⟹2x+3y-x-3y=1-z-2+2z$

$⟹x=z-1$

$⟹z=\frac{x+1}{1}$

Putting values of $z$ in (ii), we get,

$z-1+3y=2-2z$

$⟹3y=2-2z-z+1$

$3y=-3(z-1)$

$⟹y=-(z-1)=-z+1$

$\therefore z=\frac{y-1}{1}$

Hence, we have $\frac{x+1}{1}=\frac{y-1}{1}=\frac{z}{1}$

thus $\cos \alpha =\frac{a}{\left(\sqrt{(a^2+b^2+c^2)}\right)}$

$\therefore \cos \alpha =\frac{1}{\sqrt{3}}$