Vector algebra strategy — dot product vs cross product application selection

Question

Two vectors are $\vec{a} = 2\hat{i} + 3\hat{j} + \hat{k}$ and $\vec{b} = \hat{i} - 2\hat{j} + 2\hat{k}$ . Find (a) the angle between them, and (b) a vector perpendicular to both.

(CBSE 12 / JEE Main — Vectors)

Vector Product Selection

flowchart TD
    A["Vector Problem"] --> B{What do you need?}
    B -->|Angle between vectors| C["Dot Product"]
    B -->|Projection of one on another| C
    B -->|Work done (F.d)| C
    B -->|Perpendicular vector| D["Cross Product"]
    B -->|Area of parallelogram| D
    B -->|Torque (r x F)| D
    B -->|Volume of parallelepiped| E["Scalar Triple Product"]
    C --> C1["a.b = |a||b| cos theta"]
    D --> D1["a x b = |a||b| sin theta n-hat"]
    E --> E1["[a b c] = a.(b x c)"]

Solution — Step by Step

\vec{a} \cdot \vec{b} = (2)(1) + (3)(-2) + (1)(2) = 2 - 6 + 2 = -2

|\vec{a}| = \sqrt{4 + 9 + 1} = \sqrt{14}

|\vec{b}| = \sqrt{1 + 4 + 4} = \sqrt{9} = 3

\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \frac{-2}{3\sqrt{14}}

\boxed{\theta = \cos^{-1}\left(\frac{-2}{3\sqrt{14}}\right)}

The negative value tells us the angle is obtuse (between 90° and 180°).

\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 1 \\ 1 & -2 & 2 \end{vmatrix}

= \hat{i}(3 \times 2 - 1 \times (-2)) - \hat{j}(2 \times 2 - 1 \times 1) + \hat{k}(2 \times (-2) - 3 \times 1)

= \hat{i}(6 + 2) - \hat{j}(4 - 1) + \hat{k}(-4 - 3)

\boxed{\vec{a} \times \vec{b} = 8\hat{i} - 3\hat{j} - 7\hat{k}}

This vector is perpendicular to both $\vec{a}$ and $\vec{b}$ .

Check: $\vec{a} \cdot (\vec{a} \times \vec{b}) = (2)(8) + (3)(-3) + (1)(-7) = 16 - 9 - 7 = 0$ . Perpendicular.

Check: $\vec{b} \cdot (\vec{a} \times \vec{b}) = (1)(8) + (-2)(-3) + (2)(-7) = 8 + 6 - 14 = 0$ . Perpendicular.

Why This Works

The dot product captures the “alignment” between two vectors (cosine of the angle). When vectors are perpendicular, $\vec{a} \cdot \vec{b} = 0$ . The cross product produces a new vector perpendicular to both inputs — its magnitude equals the area of the parallelogram formed by the two vectors.

These are the two fundamental operations on vectors. Every vector problem ultimately reduces to one (or both) of them.

Alternative Method — Using Components Directly

For the unit vector perpendicular to both:

\hat{n} = \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} = \frac{8\hat{i} - 3\hat{j} - 7\hat{k}}{\sqrt{64 + 9 + 49}} = \frac{8\hat{i} - 3\hat{j} - 7\hat{k}}{\sqrt{122}}

For the cross product determinant, use the “cover and calculate” method: cover the first row and first column to get the $\hat{i}$ component, cover the first row and second column (with a negative sign) for $\hat{j}$ , cover the first row and third column for $\hat{k}$ . Practice this until it takes under 30 seconds — it appears in nearly every JEE paper.

Common Mistake

The order matters for cross product: $\vec{a} \times \vec{b} = -(\vec{b} \times \vec{a})$ . Reversing the order flips the direction of the result. In the determinant, if you swap the second and third rows, all signs flip. Always put the first vector in row 2 and the second in row 3.