Question

Find $\sin \, \theta $, if $\theta$ is the angle between the vectors $3 \hat{i} + \hat{j} +2\hat{k} $ and $\hat{i} + \hat{j} + 2\hat{k} $

• A. $\sqrt{\frac{5}{21}}$
• B. $ \frac{5}{\sqrt{21}}$
• C. $ \frac{4}{\sqrt{21}}$
• D. $ \frac{3}{\sqrt{21}}$

Answer 1

Answer: (A)

Let $\vec{a} = 3\hat{i} + \hat{j} + 2 \hat{k} , \vec{b} = \hat{i} + \hat{j} + 2\hat{k}$ then $ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} &\hat{j}&\hat{k} \\ 3&1&2\\ 1&1&2\end{vmatrix}$
$ =\hat{i}(2 -2) - \hat{j} (6 -2) +\hat{k} (3 -1)$
$ = - 4 \hat{j} + 2\hat{k}$
$| \vec{a} \times \vec{b} | = \sqrt{(-4)^2 + (2)^2} = \sqrt{16 + 4} = \sqrt{20} $
Since, $| \vec{a} \times \vec{b} | = | \vec{a} \times \vec{b} | \sin \theta$,
If $\theta$ is angle between $\vec{a}$ and $\vec{b}$
$\Rightarrow \sqrt{20} =\sqrt{9+1+4} \sqrt{1+1+4} \sin \theta $
$\Rightarrow \sin \theta =\frac{\sqrt{20}}{\sqrt{6} \sqrt{14}} $
$\Rightarrow \sin \theta =\sqrt{\frac{20}{6 \times14} } =\sqrt{\frac{5}{21}}$