Vectors: Application Problems (5)

Question

Find a unit vector perpendicular to both $\vec{a} = 3\hat{i} + \hat{j} + 2\hat{k}$ and $\vec{b} = 2\hat{i} - 2\hat{j} + 4\hat{k}$. Then find the sine of the angle between $\vec{a}$ and $\vec{b}$.

Solution — Step by Step

Why This Works

The cross product $\vec{a} \times \vec{b}$ has two defining properties: its direction is perpendicular to the plane containing both vectors (right-hand rule), and its magnitude is $|\vec{a}||\vec{b}|\sin\theta$. So this single operation answers both parts in one shot.

The two perpendicular unit vectors are $\pm \hat{n}$ — there are always two such vectors. The cross product picks out one direction; the negative is equally valid.

Alternative Method

For sine alone, use $\cos\theta = \vec{a}\cdot\vec{b}/(|\vec{a}||\vec{b}|) = (6 - 2 + 8)/\sqrt{14 \cdot 24} = 12/\sqrt{336}$, then $\sin\theta = \sqrt{1 - \cos^2\theta}$. More algebra, same answer — and you don’t get the unit vector for free.

Common Mistake