Vectors — H2 Maths help

Vectors underpin the 3D geometry section of H2 Mathematics (9758). You will use the scalar (dot) product to find angles between directions, the vector (cross) product to find areas and normals, and both to work with equations of lines and planes. Every technique reduces to careful manipulation of column vectors and a small set of formulae.

What you need to know

The dot product $\mathbf{a}\cdot\mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3$ , which gives $\cos\theta = \dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}$ for the acute or obtuse angle between two vectors.
The cross product $\mathbf{a}\times\mathbf{b}$ produces a vector perpendicular to both; its magnitude equals the area of the parallelogram spanned by $\mathbf{a}$ and $\mathbf{b}$ , so the area of a triangle with sides $\mathbf{a}$ and $\mathbf{b}$ is $\tfrac{1}{2}|\mathbf{a}\times\mathbf{b}|$ .
A line through point $A$ with direction $\mathbf{d}$ has vector equation $\mathbf{r} = \mathbf{a} + t\mathbf{d}$ ; the cartesian form follows by eliminating the parameter $t$ .
A plane with normal $\mathbf{n}$ through point $A$ satisfies $\mathbf{n}\cdot\mathbf{r} = \mathbf{n}\cdot\mathbf{a}$ , written in cartesian form as $n_1 x + n_2 y + n_3 z = d$ .
The distance from a point $P$ to the plane $\mathbf{n}\cdot\mathbf{r} = d$ is $\dfrac{|\mathbf{n}\cdot\mathbf{p} - d|}{|\mathbf{n}|}$ .
Two lines are skew if they are neither parallel nor intersecting; show this by attempting to solve for parameters and reaching a contradiction.

Worked example 1 — angle between two vectors

Find the angle between $\mathbf{a} = \begin{pmatrix}2\\1\\-2\end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix}1\\-2\\2\end{pmatrix}$ .

Step 1: compute the dot product.

\mathbf{a}\cdot\mathbf{b} = (2)(1) + (1)(-2) + (-2)(2) = 2 - 2 - 4 = -4.

Step 2: find the magnitudes.

|\mathbf{a}| = \sqrt{4+1+4} = 3, \qquad |\mathbf{b}| = \sqrt{1+4+4} = 3.

Step 3: apply the formula.

\cos\theta = \frac{-4}{3 \times 3} = -\frac{4}{9}.

\theta = \cos^{-1}\!\left(-\tfrac{4}{9}\right) \approx 116.4^\circ.

The angle between the two vectors is approximately $116.4^\circ$ .

Worked example 2 — area of a triangle via the cross product

Triangle $ABC$ has vertices $A(1,0,0)$ , $B(0,2,0)$ , $C(0,0,3)$ . Find its area.

Step 1: form two edge vectors from $A$ .

\overrightarrow{AB} = \begin{pmatrix}-1\\2\\0\end{pmatrix}, \qquad \overrightarrow{AC} = \begin{pmatrix}-1\\0\\3\end{pmatrix}.

Step 2: compute the cross product.

\overrightarrow{AB}\times\overrightarrow{AC} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\-1&2&0\\-1&0&3\end{vmatrix} = \begin{pmatrix}(2)(3)-(0)(0)\\(0)(-1)-(−1)(3)\\(-1)(0)-(2)(-1)\end{pmatrix} = \begin{pmatrix}6\\3\\2\end{pmatrix}.

Step 3: find the magnitude and halve it.

\left|\overrightarrow{AB}\times\overrightarrow{AC}\right| = \sqrt{36+9+4} = \sqrt{49} = 7.

\text{Area of } \triangle ABC = \tfrac{1}{2} \times 7 = 3.5 \text{ units}^2.

Worked example 3 — distance from a point to a plane

Find the perpendicular distance from $P(1,3,-1)$ to the plane $2x - y + 2z = 5$ .

The normal vector is $\mathbf{n} = \begin{pmatrix}2\\-1\\2\end{pmatrix}$ , so $|\mathbf{n}| = \sqrt{4+1+4} = 3$ .

\text{Distance} = \frac{|2(1) + (-1)(3) + 2(-1) - 5|}{3} = \frac{|2 - 3 - 2 - 5|}{3} = \frac{|-8|}{3} = \frac{8}{3} \approx 2.67 \text{ units.}

Common mistakes

Forgetting to divide by $|\mathbf{n}|$ when computing the distance to a plane — the numerator alone is not the distance.
Mixing up angle conventions: the dot-product formula gives the angle between $0^\circ$ and $180^\circ$ ; always state whether the question asks for the acute angle.
Sign errors in the cross product: use the determinant expansion carefully and check each component against the cyclic rule $\mathbf{i}\to\mathbf{j}\to\mathbf{k}$ .
Using direction vectors instead of normals (or vice versa) when writing the equation of a plane or finding the angle between a line and a plane.