Vectors: Edge Cases and Subtle Traps (7)

Question

If $\vec{a}\cdot\vec{b} = 0$ and $\vec{a}\times\vec{b} = \vec{0}$, what can we conclude about $\vec{a}$ and $\vec{b}$? Many students answer ”$\vec{a}$ is perpendicular to $\vec{b}$ and parallel to $\vec{b}$, which is impossible”. What’s the actual conclusion?

Solution — Step by Step

The conclusion: at least one of $\vec{a}, \vec{b}$ is the zero vector.

Why This Works

The dot product and cross product give complementary information about the angle. The dot product picks out $\cos\theta$, the cross product picks out $\sin\theta$. If both are zero, then for nonzero vectors, both $\cos\theta$ and $\sin\theta$ would have to be zero — impossible.

So the only escape is that one (or both) vectors has zero magnitude, in which case the products vanish for trivial reasons. This is the “edge case” examiners love testing.

Alternative Method

Directly: $|\vec{a}\times\vec{b}|^2 + (\vec{a}\cdot\vec{b})^2 = |\vec{a}|^2|\vec{b}|^2$. If both terms on the left are zero, the right side must be zero too, so $|\vec{a}||\vec{b}| = 0$, meaning at least one vector is zero. Cleaner derivation.

Common Mistake

Excluding zero vectors from consideration. JEE often plants questions where the zero vector is the only valid solution. Always ask whether $\vec{a} = \vec{0}$ is consistent with the given conditions before declaring “impossible”.