Question

The angle between the two vectors $\hat{i}+\hat{j}+\hat{k}$ and $2\hat{i}-2\hat{j}+2\hat{k}$ is equal to

• A. ${\cos }^{-1}\left(\frac{2}{3}\right)$
• B. ${\cos }^{-1}\left(\frac{1}{6}\right)$
• C. ${\cos }^{-1}\left(\frac{5}{6}\right)$
• D. ${\cos }^{-1}\left(\frac{1}{18}\right)$
• E. ${\cos }^{-1}\left(\frac{1}{3}\right)$

Answer 1

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

$\text{Step 1: Angle between two vectors}\left(\theta \right)$

Dot product between the two vectors is defined as:

$\overrightarrow{a}.\overrightarrow{b}=|\overrightarrow{a}||\overrightarrow{b}|\cos \theta$

$\Rightarrow \cos \theta =\frac{\overrightarrow{a}\cdot \overrightarrow{b}}{|\overrightarrow{a}||\overrightarrow{b}|}=\frac{(\hat{i}+\hat{j}+\hat{k})\cdot (2\hat{i}-2\hat{j}+2\hat{k})}{\sqrt{3}\times \sqrt{4+4+4}}$$=\frac{2-2+2}{\sqrt{3}\times \sqrt{12}}=\frac{1}{3}$

$\Rightarrow \theta ={\cos }^{-1}\left(\frac{1}{3}\right)$

Hence angle between vectors is ${\cos }^{-1}\left(\frac{1}{3}\right)$

Option (E) correct.