Question

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

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

Answer 1

The correct option is D ${\cos }^{-1}\left(1/9\right)$

$\overrightarrow{a}=\hat{i}+2\hat{j}+2\hat{k}$

$\overrightarrow{b}=\hat{1}-2\hat{j}+2\hat{k}$

We know,

dot product,

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

$\Rightarrow cos\theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}⟶\left(1\right)$

$\overrightarrow{a}.\overrightarrow{b}=1-4+4=1$

$\left|\overrightarrow{a}\right|=\sqrt{1^2+2^2+2^2}=3$

$\left|\overrightarrow{b}\right|=\sqrt{1^2+2^2+2^2}=3$

Using (1),

$\Rightarrow \cos \theta =\frac{1}{3.3}=\frac{1}{9}$

$\Rightarrow \cos \theta =\frac{1}{9}$

$\Rightarrow \theta ={\cos }^{-1}\frac{1}{9}$