4. Intersections of Lines and Planes

Video coming soon3 worked examples

Theory

In three dimensions, two lines need not meet even when they are not parallel — they can pass by one another at different heights. Such lines are called skew. Deciding which case applies is a standard exam question.

Two lines. Write both in parametric form using different parameters, say $s$ and $t$ , and equate the components. That gives three equations in two unknowns. Solve any two of them, then substitute into the third to check:

Third equation satisfied → the lines intersect; substitute back for the point
Third equation fails, directions not parallel → the lines are skew
Directions are scalar multiples → the lines are parallel (or identical)

A line and a plane. Substitute the parametric coordinates of the line into the plane equation. This gives a single equation in the parameter; solve it and substitute back to get the point.

Two planes. Non-parallel planes always meet in a line. Solve the two equations together, setting one variable equal to a parameter $\lambda$ , and express the other two in terms of it.

Three planes. This is a $3\times 3$ system, so use Gaussian elimination. A unique solution means the planes meet at a single point; a row of zeros means they share a whole line; an inconsistent row means there is no common point at all.

The Golden Rule: when testing two lines, always use two different parameters, and never stop after solving two equations — the third equation is the entire test. Skipping it means you cannot distinguish intersecting lines from skew ones.

⚠️ Common Examiner Traps

Using $t$ for both lines: the two lines reach the meeting point at different parameter values, so a shared letter forces a false conclusion.
Skipping the third-equation check: two equations can always be solved for two unknowns — that alone proves nothing.
Confusing skew with parallel: skew lines have non-parallel directions but still never meet. Check the directions before naming the case.
Forgetting to find the point: if the lines do intersect, the question almost always wants the coordinates, not just the fact.

Worked examples

Example 1

Show that the lines $\mathbf{r} = (1,2,3) + s(1,1,2)$ and $\mathbf{r} = (1,5,6) + t(2,-1,1)$ intersect, and find the point of intersection.

Step 1: Equate the two position vectors component by component, using different parameters:

\begin{aligned} x: &\quad 1 + s = 1 + 2t \\ y: &\quad 2 + s = 5 - t \\ z: &\quad 3 + 2s = 6 + t \end{aligned}

Step 2: Simplify the first two equations:

s = 2t, \qquad s = 3 - t

Step 3: Solve them together:

2t = 3 - t \implies 3t = 3 \implies t = 1, \quad s = 2

Step 4: Now test these values in the third equation — the step that decides the answer:

3 + 2(2) = 7 \qquad \text{and} \qquad 6 + 1 = 7 \ \checkmark

The third equation is satisfied, so the lines do intersect.

Step 5: Substitute $s = 2$ into the first line to find the point:

(1,2,3) + 2(1,1,2) = (3,\,4,\,7)

Example 2

Show that the lines $\mathbf{r} = (1,2,3) + s(2,1,-1)$ and $\mathbf{r} = (2,0,1) + t(1,-1,2)$ are skew.

Step 1: First check the directions. If $(2,1,-1)$ were a multiple of $(1,-1,2)$ , the lines would be parallel — but doubling the first component gives $2$ , while doubling the second gives $-2 \neq 1$ . They are not parallel.

Step 2: Equate the components:

\begin{aligned} x: &\quad 1 + 2s = 2 + t \\ y: &\quad 2 + s = -t \\ z: &\quad 3 - s = 1 + 2t \end{aligned}

Step 3: From the $y$ equation, $t = -2 - s$ . Substitute into the $x$ equation:

1 + 2s = 2 + (-2 - s) = -s \implies 3s = -1 \implies s = -\tfrac{1}{3}, \quad t = -\tfrac{5}{3}

Step 4: Test these values in the $z$ equation:

\text{LHS} = 3 - \left(-\tfrac{1}{3}\right) = \tfrac{10}{3}, \qquad \text{RHS} = 1 + 2\left(-\tfrac{5}{3}\right) = -\tfrac{7}{3}

Step 5: Since $\frac{10}{3} \neq -\frac{7}{3}$ , no pair of parameter values satisfies all three equations, so the lines never meet. As they are also not parallel, they are skew.

Example 3

Find the point where the line $\mathbf{r} = (1,0,2) + t(2,1,-1)$ meets the plane $3x + y - 2z = 5$ .

Step 1: Write the coordinates of a general point on the line in terms of $t$ :

x = 1 + 2t, \qquad y = t, \qquad z = 2 - t

Step 2: Substitute these into the equation of the plane:

3(1 + 2t) + t - 2(2 - t) = 5

Step 3: Expand and collect terms in $t$ :

3 + 6t + t - 4 + 2t = 5 \implies 9t - 1 = 5 \implies t = \frac{2}{3}

Step 4: Substitute back into the line to get the coordinates:

\left(1 + \tfrac{4}{3},\ \tfrac{2}{3},\ 2 - \tfrac{2}{3}\right) = \left(\tfrac{7}{3},\ \tfrac{2}{3},\ \tfrac{4}{3}\right)

Check: $3\left(\tfrac{7}{3}\right) + \tfrac{2}{3} - 2\left(\tfrac{4}{3}\right) = 7 + \tfrac{2}{3} - \tfrac{8}{3} = 5$ ✓