1. Vector & Scalar Triple Product

Video coming soon3 worked examples

Theory

The scalar (dot) product gives the angle between two vectors:

\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta \implies \cos\theta = \frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}

The vector (cross) product $\mathbf{a} \times \mathbf{b}$ is a vector perpendicular to both, computed as a determinant. The scalar triple product $\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})$ gives the volume of the parallelepiped and is zero when the vectors are coplanar.

The Golden Rule: use the scalar product for angles, the vector product when you need a perpendicular direction, and the scalar triple product to test coplanarity — a zero value means coplanar.

⚠️ Common Examiner Traps

Scalar vs vector: the dot product is a number, the cross product is a vector.
Order of the cross product: $\mathbf{a}\times\mathbf{b} = -(\mathbf{b}\times\mathbf{a})$ — the order matters.
Angle formula: the angle between vectors uses the scalar product.

Worked examples

Example 1

Find the angle between $\mathbf{a} = (1, 0, 1)$ and $\mathbf{b} = (1, 1, 0)$ .

Step 1: Compute the scalar product and the two magnitudes:

\mathbf{a}\cdot\mathbf{b} = 1 + 0 + 0 = 1, \qquad |\mathbf{a}| = \sqrt{2}, \quad |\mathbf{b}| = \sqrt{2}

Step 2: Apply the angle formula:

\cos\theta = \frac{1}{\sqrt{2}\,\sqrt{2}} = \frac{1}{2} \implies \theta = 60^\circ

Example 2

Find $\mathbf{a} \times \mathbf{b}$ for $\mathbf{a} = (1, 2, 3)$ and $\mathbf{b} = (2, 1, 0)$ .

Step 1: Write the cross product as a determinant:

\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 3 \\ 2 & 1 & 0 \end{vmatrix}

Step 2: Expand along the top row:

= \mathbf{i}(2\cdot 0 - 3\cdot 1) - \mathbf{j}(1\cdot 0 - 3\cdot 2) + \mathbf{k}(1\cdot 1 - 2\cdot 2)

Step 3: Simplify:

= (-3,\ 6,\ -3)

Example 3

Determine whether $\mathbf{a} = (1,0,1)$ , $\mathbf{b} = (1,1,0)$ and $\mathbf{c} = (2,1,1)$ are coplanar.

Step 1: First find $\mathbf{b}\times\mathbf{c}$ :

\mathbf{b}\times\mathbf{c} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & 0 \\ 2 & 1 & 1 \end{vmatrix} = (1,\ -1,\ -1)

Step 2: Compute the scalar triple product $\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})$ :

(1,0,1)\cdot(1,-1,-1) = 1 + 0 - 1 = 0

Step 3: The scalar triple product is $0$ , so the three vectors are coplanar.