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AS & A Level · Practice questions · Equations of planes

Point, normal, plane.

Six original Cambridge-style questions on equations of planes: the forms \(\mathbf{r}\cdot\mathbf{n} = p\), \(ax + by + cz = d\) and \(\mathbf{r} = \mathbf{a} + \lambda\mathbf{b} + \mu\mathbf{c}\), converting between them, and the distance from a point to a plane.

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

Find a point and a normal, then \(\mathbf{r}\cdot\mathbf{n} = \mathbf{a}\cdot\mathbf{n}\).

01
Recall
[2 marks]

Find the equation of the plane through \((3, -2, 1)\) perpendicular to \(4\mathbf{i} + \mathbf{j} - 2\mathbf{k}\), in the form \(\mathbf{r}\cdot\mathbf{n} = p\) and in Cartesian form.

  • \(p = (3)(4) + (-2)(1) + (1)(-2) = 12 - 2 - 2 = 8\). ✓
  • \(\mathbf{r}\cdot(4\mathbf{i} + \mathbf{j} - 2\mathbf{k}) = 8\), that is \(4x + y - 2z = 8\). ✓
02
Calculation
[3 marks]

Find a Cartesian equation of the plane through the points \(A(1, 1, 1)\), \(B(2, 0, 3)\) and \(C(4, 2, 0)\).

  • \(\overrightarrow{AB} = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}\), \(\overrightarrow{AC} = \begin{pmatrix} 3 \\ 1 \\ -1 \end{pmatrix}\). ✓
  • Normal \(\begin{pmatrix} p \\ q \\ s \end{pmatrix}\): \(p - q + 2s = 0\) and \(3p + q - s = 0\). Adding: \(4p + s = 0\). Take \(p = -1\), \(s = 4\); then \(q = p + 2s = 7\). So \(\mathbf{n} = \begin{pmatrix} -1 \\ 7 \\ 4 \end{pmatrix}\). ✓
  • \(\mathbf{a}\cdot\mathbf{n} = -1 + 7 + 4 = 10\): \(-x + 7y + 4z = 10\). ✓
  • Check: \(B\) gives \(-2 + 0 + 12 = 10\) and \(C\) gives \(-4 + 14 + 0 = 10\). ✓
03
Calculation
[4 marks]

The plane \(\Pi\) has equation \(\mathbf{r} = \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix} + \lambda\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} + \mu\begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}\).

(a) Find a Cartesian equation of \(\Pi\).

(b) Show that \((4, 0, 3)\) lies in \(\Pi\) and find the values of \(\lambda\) and \(\mu\) that give it.

(c) Show that \((5, 1, 2)\) does not lie in \(\Pi\).

  • (a) \(p + q = 0\) and \(p - q + 2s = 0\). So \(q = -p\) and \(2p + 2s = 0\), \(s = -p\). Take \(\mathbf{n} = \begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix}\). ✓
  • \(\mathbf{a}\cdot\mathbf{n} = 2 - 0 - 1 = 1\): \(x - y - z = 1\). ✓
  • (b) \(4 - 0 - 3 = 1\), so the point lies in \(\Pi\). Components: \(2 + \lambda + \mu = 4\), \(\lambda - \mu = 0\), \(1 + 2\mu = 3\). So \(\mu = 1\) and \(\lambda = 1\), and the first equation checks: \(2 + 1 + 1 = 4\). ✓
  • (c) \(5 - 1 - 2 = 2 \ne 1\), so \((5, 1, 2)\) is not in \(\Pi\). ✓
04
Analysis
[4 marks]

The plane \(\Pi\) has equation \(2x - 3y + z = 6\).

(a) Write \(\Pi\) in the form \(\mathbf{r} = \mathbf{a} + \lambda\mathbf{b} + \mu\mathbf{c}\).

(b) Find the exact distance of \(\Pi\) from the origin.

(c) The plane \(\Pi'\) is parallel to \(\Pi\) and passes through \((1, 1, 1)\). Find its equation and the distance between \(\Pi\) and \(\Pi'\).

  • (a) Let \(x = \lambda\), \(y = \mu\): then \(z = 6 - 2\lambda + 3\mu\), so \(\mathbf{r} = \begin{pmatrix} 0 \\ 0 \\ 6 \end{pmatrix} + \lambda\begin{pmatrix} 1 \\ 0 \\ -2 \end{pmatrix} + \mu\begin{pmatrix} 0 \\ 1 \\ 3 \end{pmatrix}\). ✓
  • Check: \((1, 0, -2)\cdot(2, -3, 1) = 0\) and \((0, 1, 3)\cdot(2, -3, 1) = 0\). ✓
  • (b) \(\dfrac{6}{\sqrt{4 + 9 + 1}} = \dfrac{6}{\sqrt{14}} = \dfrac{3\sqrt{14}}{7}\). ✓
  • (c) \(d = 2 - 3 + 1 = 0\), so \(\Pi'\) is \(2x - 3y + z = 0\), which passes through the origin. The distance between the planes is therefore \(\dfrac{3\sqrt{14}}{7} \approx 1.60\). ✓
05
Analysis
[5 marks]

The plane \(\Pi\) has equation \(\mathbf{r}\cdot(\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}) = 4\) and \(P\) is the point \((3, 4, -1)\).

(a) Find the distance of \(\Pi\) from the origin.

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

(c) Determine whether \(O\) and \(P\) are on the same side of \(\Pi\).

(d) Find the equations of the two planes parallel to \(\Pi\) that are at a distance 2 from \(\Pi\).

  • (a) \(|\mathbf{n}| = \sqrt{1 + 4 + 4} = 3\), so the distance is \(\dfrac{4}{3}\). ✓
  • (b) \(\dfrac{|3 + 8 + 2 - 4|}{3} = \dfrac{9}{3} = 3\). ✓
  • (c) \(x + 2y - 2z - 4\) is \(-4\) at \(O\) and \(+9\) at \(P\): opposite signs, so opposite sides. ✓
  • (d) A parallel plane is \(x + 2y - 2z = d\). Its distance from \(\Pi\) is \(\dfrac{|d - 4|}{3} = 2\), so \(d - 4 = \pm 6\). ✓
  • \(x + 2y - 2z = 10\) and \(x + 2y - 2z = -2\). ✓
06
Proof
[5 marks]

The plane \(\Pi\) has equation \(\mathbf{r}\cdot\mathbf{n} = p\), and \(Q\) is a point with position vector \(\mathbf{q}\).

(a) The line through \(Q\) perpendicular to \(\Pi\) is \(\mathbf{r} = \mathbf{q} + t\mathbf{n}\). Find the value of \(t\) where it meets \(\Pi\).

(b) Hence show that the perpendicular distance from \(Q\) to \(\Pi\) is \(\dfrac{|\mathbf{q}\cdot\mathbf{n} - p|}{|\mathbf{n}|}\), and that this is the formula \(\dfrac{|ax_{1} + by_{1} + cz_{1} - d|}{\sqrt{a^{2} + b^{2} + c^{2}}}\).

(c) Deduce the distance of \(\Pi\) from the origin.

  • (a) Substitute: \((\mathbf{q} + t\mathbf{n})\cdot\mathbf{n} = p\), so \(\mathbf{q}\cdot\mathbf{n} + t\,\mathbf{n}\cdot\mathbf{n} = p\) and \(t = \dfrac{p - \mathbf{q}\cdot\mathbf{n}}{|\mathbf{n}|^{2}}\). ✓
  • (b) The foot of the perpendicular is \(\mathbf{q} + t\mathbf{n}\), so the distance is \(|t\mathbf{n}| = |t|\,|\mathbf{n}| = \dfrac{|p - \mathbf{q}\cdot\mathbf{n}|}{|\mathbf{n}|^{2}}\,|\mathbf{n}| = \dfrac{|\mathbf{q}\cdot\mathbf{n} - p|}{|\mathbf{n}|}\). ✓
  • With \(\mathbf{n} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k}\), \(p = d\) and \(\mathbf{q} = x_{1}\mathbf{i} + y_{1}\mathbf{j} + z_{1}\mathbf{k}\): \(\mathbf{q}\cdot\mathbf{n} = ax_{1} + by_{1} + cz_{1}\) and \(|\mathbf{n}| = \sqrt{a^{2} + b^{2} + c^{2}}\), which gives the formula. ✓
  • (c) Put \(\mathbf{q} = \mathbf{0}\): the distance is \(\dfrac{|p|}{|\mathbf{n}|}\). ✓

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