Question

Let $\overrightarrow{a}=\hat{i}+\hat{j}-\hat{k}$ and $\overrightarrow{b}=\hat{i}-\hat{j}+\hat{k}.$
Then the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is

[2 marks]

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

Answer 1

The correct option is B ${\cos }^{-1}(-\frac{1}{3})$

$\overrightarrow{a}=\hat{i}+\hat{j}-\hat{k}$
$\Rightarrow |\overrightarrow{a}|=\sqrt{1^2+1^2+(-1{)}^2}=\sqrt{3}$
and $\overrightarrow{b}=\hat{i}-\hat{j}+\hat{k}$
$\Rightarrow |\overrightarrow{b}|=\sqrt{1^2+(-1{)}^2+1^2}=\sqrt{3}$
Now, $\overrightarrow{a}\cdot \overrightarrow{b}=(\hat{i}+\hat{j}-\hat{k})\cdot (\hat{i}-\hat{j}+\hat{k})$
$\Rightarrow \overrightarrow{a}\cdot \overrightarrow{b}=1-1-1=-1$

Let $\theta$ be the angle between the vectors $\overrightarrow{a}$ and $\overrightarrow{b}.$
Then, $\cos \theta =\frac{\overrightarrow{a}\cdot \overrightarrow{b}}{|\overrightarrow{a}||\overrightarrow{b}|}$
$\Rightarrow \cos \theta =\frac{-1}{\sqrt{3}\sqrt{3}}$
$\Rightarrow \theta ={\cos }^{-1}(-\frac{1}{3})$