Vectors: PYQ Walkthrough (14)

Question

If $\vec{a} = 2\hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + 2\hat{k}$, find (a) the angle between $\vec{a}$ and $\vec{b}$, (b) the projection of $\vec{a}$ on $\vec{b}$, and (c) a unit vector perpendicular to both.

This is the standard JEE Main vectors three-in-one. The same setup is asked yearly with different numbers.

Solution — Step by Step

Final answers: (a) $\cos^{-1}(-1/6)$, (b) $-1/\sqrt{6}$, (c) $(\hat{i} - 5\hat{j} - 3\hat{k})/\sqrt{35}$.

Why This Works

The dot product extracts how much one vector “points along” another, scaled by magnitudes. Negative dot product means obtuse angle. The projection is the component of $\vec{a}$ along the direction of $\vec{b}$ — a scalar that can be negative.

The cross product gives a vector perpendicular to both, with magnitude equal to the area of the parallelogram spanned by them. Dividing by its magnitude gives the unit perpendicular.

Alternative Method

For the angle, you can use $\sin\theta = |\vec{a}\times\vec{b}|/(|\vec{a}||\vec{b}|) = \sqrt{35}/6$. Combined with the negative cosine, this places $\theta$ in the second quadrant. Same answer, useful as a cross-check.

Common Mistake

Mixing up the projection of $\vec{a}$ on $\vec{b}$ versus $\vec{b}$ on $\vec{a}$. They are different in general:

$\text{proj}_{\vec{b}}\vec{a} = \vec{a}\cdot\vec{b}/|\vec{b}|$, $\text{proj}_{\vec{a}}\vec{b} = \vec{a}\cdot\vec{b}/|\vec{a}|$.

Same dot product, different denominators.