Question

If $\overrightarrow{A}=\hat{i}+\hat{j}+\hat{k}$ and $\overrightarrow{B}=2\hat{i}-\hat{j}+3\hat{k}$. Find the angle between $\overrightarrow{A}$ and $\overrightarrow{B}$.

• A. $\cos -1\left(\frac{4}{\sqrt{42}}\right)$
• B. $\cos -1\left(\frac{2}{\sqrt{14}}\right)$
• C. $\cos -1\left(\frac{2}{\sqrt{42}}\right)$
• D. $\cos -1\left(\frac{4}{\sqrt{3}}\right)$

Answer 1

The correct option is A $\cos -1\left(\frac{4}{\sqrt{42}}\right)$

Given, $\overrightarrow{A}=\hat{i}+\hat{j}+\hat{k}$ and $\overrightarrow{B}=2\hat{i}-\hat{j}+3\hat{k}$
We know that,
$\cos \theta =\frac{\overrightarrow{A}.\overrightarrow{B}}{|\overrightarrow{A}||\overrightarrow{B}|}$
$\Rightarrow \cos \theta =\frac{(\hat{i}+\hat{j}+\hat{k}).(2\hat{i}-\hat{j}+3\hat{k})}{\sqrt{1^2+1^2+1^2}\times \sqrt{2^2+(-1{)}^2+3^2}}$
$\Rightarrow \cos \theta =\frac{2-1+3}{\sqrt{\left(3\right)\left(14\right)}}=\frac{4}{\sqrt{42}}$
$\Rightarrow \theta =\cos -1\left(\frac{4}{\sqrt{42}}\right)$