AS & A Level · Practice questions · Lines and planes: intersections and angles

Substitute, solve, measure.

Six original Cambridge-style questions on lines and planes: deciding whether a line lies in a plane, finding where a line meets a plane, the foot of the perpendicular and the distance from a point, angles with sine and cosine, the line of intersection of two planes and the reflection of a point.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Write the general point, then use \(\mathbf{n}\).

Intersection: write the line's general point in terms of the parameter, substitute into the plane, solve, substitute back and check.
Parallel or in: if \(\mathbf{b} \cdot \mathbf{n} = 0\), test one point of the line in the plane.
Foot, distance, image: use the line \(\mathbf{a} + t\mathbf{n}\); the foot is at \(t\), the image at \(2t\), the distance is \(|t||\mathbf{n}|\).
Angles: sine for a line and a plane, cosine of the normals for two planes, and a modulus for the acute angle.
Line of intersection: direction \(\mathbf{n}_{1} \times \mathbf{n}_{2}\), and a common point found by setting one coordinate to zero.

Recall

[2 marks]

Show that the line \(\mathbf{r} = \begin{pmatrix}3\\1\\-2\end{pmatrix} + t\begin{pmatrix}1\\-1\\1\end{pmatrix}\) lies in the plane \(x + 2y + z = 3\).

\(\mathbf{b} \cdot \mathbf{n} = 1 - 2 + 1 = 0\), so the line is parallel to the plane. ✓
The point \((3, 1, -2)\) gives \(3 + 2 - 2 = 3\), so it lies on the plane; hence the whole line lies in the plane. ✓

Calculation

[3 marks]

The line \(l\) has equation \(\dfrac{x - 1}{2} = y + 2 = \dfrac{z - 3}{-2}\). Find the coordinates of the point where \(l\) meets the plane \(\mathbf{r} \cdot \begin{pmatrix}1\\-1\\2\end{pmatrix} = 15\).

General point of \(l\): \((1 + 2t,\ -2 + t,\ 3 - 2t)\). ✓
\((1 + 2t) - (-2 + t) + 2(3 - 2t) = 15\), so \(9 - 3t = 15\) and \(t = -2\). ✓
The point is \((-3, -4, 7)\). Check: \(-3 + 4 + 14 = 15\). ✓

Calculation

[4 marks]

The plane \(\Pi\) has equation \(2x + 3y - 6z = -8\) and \(A\) is the point \((1, 5, -4)\).

(a) Find the coordinates of the foot of the perpendicular from \(A\) to \(\Pi\).

(b) Find the perpendicular distance from \(A\) to \(\Pi\).

(a) Normal line: \(\mathbf{r} = \begin{pmatrix}1\\5\\-4\end{pmatrix} + t\begin{pmatrix}2\\3\\-6\end{pmatrix}\). ✓
\(2(1 + 2t) + 3(5 + 3t) - 6(-4 - 6t) = -8\), so \(41 + 49t = -8\) and \(t = -1\). ✓
Foot \((-1, 2, 2)\). Check: \(-2 + 6 - 12 = -8\). ✓
(b) Distance \(= |t||\mathbf{n}| = 1 \times \sqrt{4 + 9 + 36} = 7\). (Formula: \(\dfrac{|41 + 8|}{7} = 7\).) ✓

Analysis

[5 marks]

The points \(A(1, 0, 2)\) and \(B(3, 1, 0)\) lie on the line \(l\). The plane \(\Pi\) has equation \(4x - 4y + 7z = 28\).

(a) Find the acute angle between \(l\) and \(\Pi\).

(b) Find the coordinates of the point where \(l\) meets \(\Pi\).

(a) \(\overrightarrow{AB} = \begin{pmatrix}2\\1\\-2\end{pmatrix}\), \(|\overrightarrow{AB}| = 3\), \(\mathbf{n} = \begin{pmatrix}4\\-4\\7\end{pmatrix}\), \(|\mathbf{n}| = 9\). ✓
\(\overrightarrow{AB} \cdot \mathbf{n} = 8 - 4 - 14 = -10\), so \(\sin\theta = \dfrac{10}{27}\) and \(\theta = 21.7^\circ\). ✓
(b) \(l: \mathbf{r} = \begin{pmatrix}1\\0\\2\end{pmatrix} + t\begin{pmatrix}2\\1\\-2\end{pmatrix}\), general point \((1 + 2t,\ t,\ 2 - 2t)\). ✓
\(4(1 + 2t) - 4t + 7(2 - 2t) = 28\), so \(18 - 10t = 28\) and \(t = -1\). ✓
The point is \((-1, -1, 4)\). Check: \(-4 + 4 + 28 = 28\). ✓

Analysis

[5 marks]

The planes \(\Pi_{1}\) and \(\Pi_{2}\) have equations \(x + 2y - z = 1\) and \(2x - y + z = 7\).

(a) Find a vector equation for the line of intersection of \(\Pi_{1}\) and \(\Pi_{2}\).

(b) Find the acute angle between \(\Pi_{1}\) and \(\Pi_{2}\).

(a) \(\begin{pmatrix}1\\2\\-1\end{pmatrix} \times \begin{pmatrix}2\\-1\\1\end{pmatrix} = \begin{pmatrix}1\\-3\\-5\end{pmatrix}\). ✓
Put \(z = 0\): \(x + 2y = 1\) and \(2x - y = 7\), so \(x = 3,\ y = -1\). ✓
\(\mathbf{r} = \begin{pmatrix}3\\-1\\0\end{pmatrix} + \lambda\begin{pmatrix}1\\-3\\-5\end{pmatrix}\). ✓
(b) \(\mathbf{n}_{1} \cdot \mathbf{n}_{2} = 2 - 2 - 1 = -1\), \(|\mathbf{n}_{1}| = |\mathbf{n}_{2}| = \sqrt{6}\). ✓
\(\cos\theta = \dfrac{1}{6}\), so \(\theta = 80.4^\circ\). ✓

Analysis

[5 marks]

The point \(A\) is \((2, 3, -1)\) and the plane \(\Pi\) has equation \(x - 2y + 2z = 3\).

(a) Find the foot \(N\) of the perpendicular from \(A\) to \(\Pi\), and the distance \(AN\).

(b) Find the reflection \(A'\) of \(A\) in \(\Pi\), and show that \(A'\) is the same distance from \(\Pi\) as \(A\).

(a) Normal line: \(\mathbf{r} = \begin{pmatrix}2\\3\\-1\end{pmatrix} + t\begin{pmatrix}1\\-2\\2\end{pmatrix}\). ✓
\((2 + t) - 2(3 - 2t) + 2(-1 + 2t) = 3\), so \(-6 + 9t = 3\) and \(t = 1\): \(N = (3, 1, 1)\). ✓
\(AN = |t||\mathbf{n}| = 1 \times 3 = 3\). ✓
(b) \(\mathbf{a}' = \mathbf{a} + 2t\mathbf{n} = (4, -1, 3)\). ✓
Distance of \(A'\): \(\dfrac{|4 + 2 + 6 - 3|}{3} = \dfrac{9}{3} = 3\), the same as \(AN\). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.