2. Equations of Lines

Video coming soon2 worked examples

Theory

A line in three dimensions needs a point $\mathbf{a}$ on it and a direction vector $\mathbf{d}$ . It can be written in three equivalent ways:

\text{Vector: } \mathbf{r} = \mathbf{a} + t\mathbf{d} \qquad \text{Symmetric: } \frac{x - a_1}{d_1} = \frac{y - a_2}{d_2} = \frac{z - a_3}{d_3}

The angle between two lines is the angle between their direction vectors.

The Golden Rule: a line is a point plus a direction. Given two points, subtract to get the direction, then convert freely between vector, parametric and symmetric forms.

⚠️ Common Examiner Traps

Direction from two points: subtract the position vectors to get $\mathbf{d}$ .
Symmetric denominators: these are the components of the direction vector.
Angle between lines: use the direction vectors, not the points.

Worked examples

Example 1

Find the vector and symmetric equations of the line through $A(1, 2, 3)$ and $B(4, 0, 5)$ .

Step 1: The direction vector is $\mathbf{d} = B - A$ :

\mathbf{d} = (4-1,\ 0-2,\ 5-3) = (3,\ -2,\ 2)

Step 2: Vector equation, using point $A$ :

\mathbf{r} = (1, 2, 3) + t(3, -2, 2)

Step 3: Symmetric form:

\frac{x-1}{3} = \frac{y-2}{-2} = \frac{z-3}{2}

Example 2

Find the acute angle between the lines with direction vectors $\mathbf{d}_1 = (1, 0, 1)$ and $\mathbf{d}_2 = (1, 1, 0)$ .

Step 1: Use the scalar product of the direction vectors:

\mathbf{d}_1\cdot\mathbf{d}_2 = 1, \qquad |\mathbf{d}_1| = |\mathbf{d}_2| = \sqrt{2}

Step 2: Apply the angle formula:

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