Question

Consider the lines:
$L_1:\frac{x+1}{3}=\frac{y+2}{1}=\frac{z+1}{2}$ and $L_2:\frac{x-2}{1}=\frac{y+2}{2}=\frac{z-3}{3}$.

The unit vector perpendicular to both $L_1$ and $L_2$ is

• A. $\frac{1}{\sqrt{99}}(-\hat{i}+7\hat{j}+7\hat{k})$
• B. $\frac{1}{5\sqrt{3}}(-\hat{i}-7\hat{j}+5\hat{k})$
• C. $\frac{1}{5\sqrt{3}}(-\hat{i}+7\hat{j}+5\hat{k})$
• D. $\frac{1}{\sqrt{99}}(7\hat{i}-7\hat{j}-\hat{k})$

Answer 1

The correct option is B $\frac{1}{5\sqrt{3}}(-\hat{i}-7\hat{j}+5\hat{k})$

Vector perpendicular to both $L_1$ and $L_2$ is given by, $n_1\times n_2$
where $n_1=3i+j+2k,n_2=i+2j+3k$

$\Rightarrow n_1\times n_2=\left|\begin{array}{ccc}i& j& k\\ 3& 1& 2\\ 1& 2& 3\end{array}\right|=-i-7j+5k$

Hence required unit vector is, $\frac{n_1\times n_2}{|n_1\times n_2|}=\frac{1}{5\sqrt{3}}(-i-7j+5k)$