Question

The unit vector perpendicular to the plane containing $\overrightarrow{A},\overrightarrow{B}$ such that $\overrightarrow{A}=4\hat{i}-\hat{j}-\hat{k}$ and $\overrightarrow{B}=4\hat{i}+\hat{j}-4\hat{k}$ is:

• A. $\frac{5\hat{i}-5\hat{j}+5\hat{k}}{\sqrt{3}}$
• B. $\frac{5\hat{i}+12\hat{j}+8\hat{k}}{\sqrt{233}}$
• C. $\frac{\hat{i}+3\hat{j}+\hat{k}}{11}$
• D. $5\hat{i}-5\hat{j}+5\hat{k}$

Answer 1

Answer: (B)

$\frac{5\hat{i}+12\hat{j}+8\hat{k}}{\sqrt{233}}$

$|\overrightarrow{A}|=\sqrt{4^2+(-1{)}^2+(-1{)}^2}=\sqrt{18}$
$|\overrightarrow{B}|=\sqrt{4^2+(1{)}^2+(-4{)}^2}=\sqrt{33}$
$\overrightarrow{A}\times \overrightarrow{B}=(4+1)\hat{i}+(-4+16)\hat{j}+(4+4)\hat{k}=5\hat{i}+12\hat{j}+8\hat{k}$
Unit vector in the direction of $\overrightarrow{A}\times \overrightarrow{B}$ is $\frac{\overrightarrow{A}\times \overrightarrow{B}}{|\overrightarrow{A}\times \overrightarrow{B}|}=\frac{5\hat{i}+12\hat{j}+8\hat{k}}{\sqrt{5^2+{12}^2+8^2}}=\frac{5\hat{i}+12\hat{j}+8\hat{k}}{\sqrt{233}}$