Question

The angle between the vectors $\overrightarrow{a}=i+2j+2k\text{and}\overrightarrow{b}=i-2j+2k$ is?

• A. ${\sin }^{-1}\left(\frac{1}{9}\right)$
• B. ${\cos }^{-1}\left(\frac{8}{9}\right)$
• C. ${\sin }^{-1}\left(\frac{8}{9}\right)$
• D. ${\cos }^{-1}\left(\frac{1}{9}\right)$

Answer 1

The correct option is D

${\cos }^{-1}\left(\frac{1}{9}\right)$

Explanation for the correct option:

Find the angle between two vectors :

Given,$\overrightarrow{a}=i+2j+2k\text{and}\overrightarrow{b}=i-2j+2k$

We know that

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

Therefore,

$$\begin{gathered} \cos \theta =\frac{\left(i+2j+2k\right).\left(i-2j+2k\right)}{\left(\sqrt{1^2+2^2+2^2}\right)\left(\sqrt{1^2+{\left(-2\right)}^2+2^2}\right)}\left[\because \left|\overrightarrow{a}\right|=\sqrt{{a_1}^2+{a_2}^2+{a_3}^2}\right] \\ \Rightarrow \cos \theta =\frac{1-4+4}{3.3} \\ \Rightarrow \cos \theta =\frac{1}{9} \\ \Rightarrow \theta ={\cos }^{-1}\left(\frac{1}{9}\right) \end{gathered}$$

Hence, option (D) is the correct answer.