If $\theta $ is the angle between the vectors $2\hat{i}-2\hat{j}+4\hat{k}$ and $3\hat{i}+\hat{j}+2\hat{k}$, then value of $sin \theta $ is :

$\frac{2}{\sqrt{7}}$

The correct answer is Option (2) → $\frac{2}{\sqrt{7}}$

$\vec a=2\hat{i}-2\hat{j}+4\hat{k}$ $|\vec a|=\sqrt{2^2+(-2)^2+4^2}=2\sqrt{6}$

$\vec b=3\hat{i}+\hat{j}+2\hat{k}$ $|\vec b|=\sqrt{14}$

$\vec a.\vec b=6-2+8=|\vec a||\vec b|\cos θ$

so $\cos θ=\frac{12}{2\sqrt{6}\sqrt{14}}=\frac{6}{\sqrt{6}\sqrt{14}}=\sqrt{\frac{6}{14}}=\sqrt{\frac{3}{7}}$

$\sin θ=\sqrt{1-\frac{3}{7}}=\sqrt{\frac{4}{7}}=\frac{2}{\sqrt{7}}$