Vectors: Numerical Problems Set (11)

Question

If $\vec{a} = 2\vec{i} + 3\vec{j} - \vec{k}$ and $\vec{b} = \vec{i} - 2\vec{j} + 3\vec{k}$, find (a) $\vec{a} \cdot \vec{b}$, (b) $\vec{a} \times \vec{b}$, and (c) the angle between $\vec{a}$ and $\vec{b}$.

Solution — Step by Step

Why This Works

The dot product encodes both magnitudes and the cosine of the angle between vectors: $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta$. The cross product gives a vector perpendicular to both, with magnitude $|\vec{a}| |\vec{b}| \sin\theta$.

A negative dot product immediately tells us the angle is obtuse (between $90°$ and $180°$). The magnitudes happened to be equal here ($\sqrt{14}$ each), making the cosine value clean.

Alternative Method

For the angle, you can also use the cross product magnitude: $|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta$. Here $|\vec{a} \times \vec{b}| = \sqrt{49 + 49 + 49} = 7\sqrt{3}$. So $\sin\theta = 7\sqrt{3}/14 = \sqrt{3}/2$, giving $\theta = 60°$ or $120°$. Combine with the negative cosine to confirm $\theta = 120°$.

Common Mistake

Students sometimes drop minus signs in cross product cofactor expansion (the middle term has a minus sign by definition). Write the determinant out fully each time — it’s safer than memorising the formula.