Question

Find $\sin θ$, if $θ$ 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 $\vec{a}×\vec{b}=â\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 1& 2\\ 1& 1& 2\end{array}â$
$=\hat{i}(2−2)−\hat{j}(6−2)+\hat{k}(3−1)$
$=−4\hat{j}+2\hat{k}$
$âˆ\pounds \vec{a}×\vec{b}âˆ\pounds =\sqrt{(−4{)}^2+(2{)}^2}=\sqrt{16+4}=\sqrt{20}$
Since, $âˆ\pounds \vec{a}×\vec{b}âˆ\pounds =âˆ\pounds \vec{a}×\vec{b}âˆ\pounds \sin θ$,
If $θ$ is angle between $\vec{a}$ and $\vec{b}$
$⇒\sqrt{20}=\sqrt{9+1+4}\sqrt{1+1+4}\sin θ$
$⇒\sin θ=\frac{\sqrt{20}}{\sqrt{6}\sqrt{14}}$
$⇒\sin θ=\sqrt{\frac{20}{6×14}}=\sqrt{\frac{5}{21}}$