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

Answer 1

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

The direction vector of the line of the intersection of the planes $2x+3y+z=1$ and $x+3y+2x=2$ is given by

$n_1\times n_2$

$=(2i+3j+k)\times (i+3j+2k)$
$=3i-3j+3k$

Hence the unit vector along the direction of the line will be

$-\frac{i-j+k}{\sqrt{3}}$

Thus

$cos\alpha =\frac{1}{\sqrt{3}}$ where $\alpha$ is the angle that the line makes with x axis.