3.16 The vector productIB Maths: Analysis and Approaches HL: Revision notes

The vector product (or cross product) of two vectors $\mathbf{v}$ and $\mathbf{w}$ is a vector, defined by $\mathbf{v}\times\mathbf{w} = |\mathbf{v}||\mathbf{w}|\sin\theta\,\mathbf{n},$ where $\theta$ is the angle between $\mathbf{v}$ and $\mathbf{w}$ ($0 \le \theta \le \pi$) and $\mathbf{n}$ is a unit vector perpendicular to both $\mathbf{v}$ and $\mathbf{w}$. The direction of $\mathbf{n}$ is given by the right-hand screw rule: turn a screw from $\mathbf{v}$ towards $\mathbf{w}$ and it moves in the direction of $\mathbf{n}$.

So $\mathbf{v}\times\mathbf{w}$ is perpendicular to both vectors, which is why it is so useful for finding normals to planes. Compare the scalar product $\mathbf{v}\cdot\mathbf{w} = |\mathbf{v}||\mathbf{w}|\cos\theta$, which is a number.

The formula booklet gives $\mathbf{v}\times\mathbf{w} = \begin{pmatrix} v_2w_3 - v_3w_2 \\ v_3w_1 - v_1w_3 \\ v_1w_2 - v_2w_1 \end{pmatrix}.$ Many students prefer the determinant layout: put $\mathbf{i}, \mathbf{j}, \mathbf{k}$ in the top row, $\mathbf{v}$ in the second and $\mathbf{w}$ in the third, then expand, remembering the minus sign on the $\mathbf{j}$ term.

Example: $\mathbf{u} = 2\mathbf{i} - \mathbf{j} + 3\mathbf{k}$, $\mathbf{v} = \mathbf{i} + 4\mathbf{j} - 2\mathbf{k}$ gives $\mathbf{u}\times\mathbf{v} = (2 - 12)\mathbf{i} + (3 + 4)\mathbf{j} + (8 + 1)\mathbf{k} = -10\mathbf{i} + 7\mathbf{j} + 9\mathbf{k}.$ Check with the scalar product: $(-10)(2) + (7)(-1) + (9)(3) = 0$, so the answer is perpendicular to $\mathbf{u}$, as it must be.

• Anti-commutative: $\mathbf{v}\times\mathbf{w} = -\mathbf{w}\times\mathbf{v}$. Order matters.
• Distributive: $\mathbf{u}\times(\mathbf{v}+\mathbf{w}) = \mathbf{u}\times\mathbf{v} + \mathbf{u}\times\mathbf{w}$.
• Scalars come out: $(k\mathbf{v})\times\mathbf{w} = k(\mathbf{v}\times\mathbf{w})$.
• $\mathbf{v}\times\mathbf{v} = \mathbf{0}$, because $\sin 0 = 0$.
• For non-zero vectors, $\mathbf{v}\times\mathbf{w} = \mathbf{0}$ exactly when $\mathbf{v}$ and $\mathbf{w}$ are parallel.

Example: $(\mathbf{a}+\mathbf{b})\times(\mathbf{a}-\mathbf{b}) = \mathbf{a}\times\mathbf{a} - \mathbf{a}\times\mathbf{b} + \mathbf{b}\times\mathbf{a} - \mathbf{b}\times\mathbf{b} = -2(\mathbf{a}\times\mathbf{b})$.

$|\mathbf{v}\times\mathbf{w}| = |\mathbf{v}||\mathbf{w}|\sin\theta$ is the area of the parallelogram with sides $\mathbf{v}$ and $\mathbf{w}$ (base $|\mathbf{v}|$, height $|\mathbf{w}|\sin\theta$). The triangle with the same two sides has half this area: $\text{area of triangle ABC} = \tfrac{1}{2}|\overrightarrow{AB}\times\overrightarrow{AC}|.$ Because area $= \frac{1}{2}\times$ base $\times$ height, you can also find the perpendicular distance from C to the line AB: $h = \dfrac{|\overrightarrow{AB}\times\overrightarrow{AC}|}{|\overrightarrow{AB}|}$.

Example: for A$(1,0,2)$, B$(5,4,4)$, C$(-2,4,3)$, $\overrightarrow{AB}\times\overrightarrow{AC} = -4\mathbf{i} - 10\mathbf{j} + 28\mathbf{k}$ with magnitude 30, so the triangle has area 15 and C is $30 \div 6 = 5$ units from AB.

Since $|\mathbf{v}\times\mathbf{w}| = |\mathbf{v}||\mathbf{w}|\sin\theta$ and $\mathbf{v}\cdot\mathbf{w} = |\mathbf{v}||\mathbf{w}|\cos\theta$, dividing gives $\tan\theta = \frac{|\mathbf{v}\times\mathbf{w}|}{\mathbf{v}\cdot\mathbf{w}}.$ For example, if $\mathbf{a}\times\mathbf{b} = 2\mathbf{i} - 3\mathbf{j} + 6\mathbf{k}$ and $\mathbf{a}\cdot\mathbf{b} = 12$, then $|\mathbf{a}\times\mathbf{b}| = 7$ and $\tan\theta = \frac{7}{12}$.

To find an angle from vectors alone, the scalar product is safer: $\sin\theta$ cannot tell an acute angle from its obtuse partner, because $|\mathbf{v}\times\mathbf{w}| \ge 0$ always.