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 $\cos \alpha =$

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

Answer 1

Answer: (A)

$\frac{1}{\sqrt{3}}$

$2x+3y+z=1$

$x+3y+2z=2$

$\hat{n}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 3& 1\\ 1& 3& 2\end{array}\right|$

$=\hat{i}(6-3)-\hat{j}(4-1)+\hat{k}(6-3)$

$\hat{n}=3\hat{i}-3\hat{j}+3\hat{k}$

$\therefore \overrightarrow{a}=\frac{3\hat{i}-3\hat{j}+3\hat{k}}{\sqrt{9+9+9}}$

$=\frac{1}{\sqrt{3}}\hat{i}-\frac{1}{\sqrt{3}}\hat{j}+\frac{1}{\sqrt{3}}\hat{k}$

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