Question

A unit vector perpendicular to the plane of $a=2\hat{i}-6\hat{j}-3\hat{k},b=4\hat{i}+3\hat{j}-\hat{k}$ is

• A. $\frac{4\hat{i}+3\hat{j}-\hat{k}}{\sqrt{26}}$
• B. $\frac{2\hat{i}-6\hat{j}-3\hat{k}}{7}$
• C. $\frac{3\hat{i}-2\hat{j}+6\hat{k}}{7}$
• D. $\frac{2\hat{i}-3\hat{j}-6\hat{k}}{7}$

Answer 1

The correct option is C $\frac{3\hat{i}-2\hat{j}+6\hat{k}}{7}$

$a\times b=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& -6& -3\\ 4& 3& -1\end{array}\right|=\hat{i}(6+9)-\hat{j}(-2+12)+\hat{k}(6+24)=15\hat{i}-10\hat{j}+30\hat{k}$
$|a\times b|=\sqrt{225+100+900}=35$
Unit vector normal to the plane = $\frac{15\hat{i}-10\hat{j}+30\hat{k}}{35}=\frac{3\hat{i}-2\hat{j}+6\hat{k}}{7}$