A-Level Further Maths 9231 / Topic 1.6 / AS & A Level
Where a line meets a plane.
Substitute the line into the plane and one equation in the parameter tells you everything: one point of intersection, no point at all, or every point. The same idea gives the foot of a perpendicular, the distance from a point to a plane and reflections, while the scalar and vector products give the angles and the line where two planes cross.
For a line \(\mathbf{r} = \mathbf{a} + t\mathbf{b}\) and a plane \(\mathbf{r} \cdot \mathbf{n} = d\): if \(\mathbf{b} \cdot \mathbf{n} \ne 0\) they meet at exactly one point; if \(\mathbf{b} \cdot \mathbf{n} = 0\) the line is parallel to the plane, and it lies IN the plane when \(\mathbf{a} \cdot \mathbf{n} = d\). The angle \(\theta\) between the line and the plane satisfies \(\sin\theta = \dfrac{|\mathbf{b} \cdot \mathbf{n}|}{|\mathbf{b}||\mathbf{n}|}\), and two planes meet at the acute angle between their normals along a line with direction \(\mathbf{n}_{1} \times \mathbf{n}_{2}\).
Fig. 1: a line with direction b meets the plane Π at P. The angle θ between the line and the plane is measured from the line's projection in the plane; the angle between the line and the normal n is 90° - θ
Section 01
In it, parallel to it, or through it.
To see how the line \(\mathbf{r} = \mathbf{a} + t\mathbf{b}\) meets the plane \(\mathbf{r} \cdot \mathbf{n} = d\), write the general point of the line as \((x, y, z)\) in terms of \(t\) and substitute it into the equation of the plane. You get a linear equation in \(t\), and there are only three things it can do.
Find the point where the line \(\mathbf{r} = \begin{pmatrix}1\\-2\\3\end{pmatrix} + t\begin{pmatrix}2\\1\\-1\end{pmatrix}\) meets the plane \(\mathbf{r} \cdot \begin{pmatrix}3\\-1\\2\end{pmatrix} = 17\).
General point of the line: \((1 + 2t,\ -2 + t,\ 3 - t)\).
\(3 + 6t + 2 - t + 6 - 2t = 17\), so \(11 + 3t = 17\) and \(t = 2\).
Put \(t = 2\) back into the line: \((5,\ 0,\ 1)\). Check: \(15 - 0 + 2 = 17\).
Answer: \((5, 0, 1)\)
Worked example Parallel or in the plane?
The plane \(\Pi\) has equation \(x + 2y - z = 4\). Decide how each line is related to \(\Pi\). (a) \(\mathbf{r} = \begin{pmatrix}2\\1\\0\end{pmatrix} + s\begin{pmatrix}1\\0\\1\end{pmatrix}\) (b) \(\mathbf{r} = \begin{pmatrix}1\\1\\1\end{pmatrix} + t\begin{pmatrix}3\\-1\\1\end{pmatrix}\)
Test \((1, 1, 1)\): \(1 + 2 - 1 = 2 \ne 4\), so this line never meets \(\Pi\).
Answer: (a) lies in \(\Pi\) (b) parallel to \(\Pi\), not in it
Your turn
Find where the line \(\mathbf{r} = \begin{pmatrix}0\\1\\2\end{pmatrix} + t\begin{pmatrix}1\\1\\1\end{pmatrix}\) meets the plane \(x + y + z = 12\).
\(t + (1 + t) + (2 + t) = 12\), so \(3 + 3t = 12\), \(t = 3\): the point \((3, 4, 5)\).
Examiner trap
\(\mathbf{b} \cdot \mathbf{n} = 0\) does NOT mean the line is in the plane. It only says the line runs parallel to the plane; you must still test one point of the line in the equation of the plane.
Section 02
The foot, the distance and the mirror image.
The shortest route from a point \(A\) to a plane runs along the normal. So the line through \(A\) with direction \(\mathbf{n}\), \(\mathbf{r} = \mathbf{a} + t\mathbf{n}\), meets the plane at the foot of the perpendicular \(N\), and \(AN\) is the perpendicular distance.
Foot of the perpendicular: substitute \(\mathbf{a} + t\mathbf{n}\) into the plane, solve for \(t\), and put \(t\) back.
Distance from \((x_{1}, y_{1}, z_{1})\) to \(ax + by + cz = d\):
\[\frac{|ax_{1} + by_{1} + cz_{1} - d|}{\sqrt{a^{2} + b^{2} + c^{2}}}\]
It also equals \(|t||\mathbf{n}|\) from the foot calculation.
Reflection of \(A\) in the plane: \(N\) is the midpoint of \(AA'\), so \(\mathbf{a}' = \mathbf{a} + 2t\mathbf{n}\) with the same \(t\).
Worked example Foot and distance
Find the foot of the perpendicular from \(A(5, 1, 7)\) to the plane \(2x - y + 2z = 5\), and the perpendicular distance from \(A\) to the plane.
Line through \(A\) along the normal: \(\mathbf{r} = \begin{pmatrix}5\\1\\7\end{pmatrix} + t\begin{pmatrix}2\\-1\\2\end{pmatrix}\).
In the distance formula the plane must be written as \(ax + by + cz = d\) and you subtract \(d\). With \(2x - y + 2z = 5\) and \(A(5, 1, 7)\) the top is \(|23 - 5|\), not \(|23 + 5|\). And divide by \(|\mathbf{n}|\), not by \(|\mathbf{n}|^{2}\).
Stage 1 · Learn
Check your understanding
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The line \(\mathbf{r} = \begin{pmatrix}1\\0\\0\end{pmatrix} + t\begin{pmatrix}1\\1\\1\end{pmatrix}\) and the plane \(x - 2y + z = 5\). Which is true?
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A line makes an angle of \(70^\circ\) with the normal to a plane. What is the angle between the line and the plane?
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Find the perpendicular distance from \((2, 1, -1)\) to the plane \(2x + y + 2z = 12\).
Quick check+10 XP
Which vector is a direction of the line of intersection of \(x + 2y - z = 4\) and \(y + z = 1\)?
Section 03
Angles and the line where planes cross.
Angles come from the scalar product, but you must choose the right pair of vectors. For two planes, use their normals and take the acute angle. For a line and a plane, the normal makes the angle \(90^\circ - \theta\) with the line (Fig. 1), which turns cosine into sine.
Line and plane: \(\sin\theta = \dfrac{|\mathbf{b} \cdot \mathbf{n}|}{|\mathbf{b}||\mathbf{n}|}\).
Two planes: \(\cos\theta = \dfrac{|\mathbf{n}_{1} \cdot \mathbf{n}_{2}|}{|\mathbf{n}_{1}||\mathbf{n}_{2}|}\), the acute angle.
Line of intersection: it lies in both planes, so it is perpendicular to both normals and has direction \(\mathbf{n}_{1} \times \mathbf{n}_{2}\). For a point on it, set one coordinate (often \(z = 0\)) and solve the two plane equations. Alternatively, set \(z = \lambda\) and solve for \(x\) and \(y\) in terms of \(\lambda\).
Worked example Angle between a line and a plane
Find the angle between the line \(\mathbf{r} = \begin{pmatrix}1\\0\\2\end{pmatrix} + t\begin{pmatrix}1\\2\\2\end{pmatrix}\) and the plane \(2x - y + 2z = 7\).
\(\theta = 26.4^\circ\). (The angle with the normal is \(63.6^\circ\), its complement.)
Answer: \(26.4^\circ\) (to 1 d.p.)
Worked example Line of intersection and angle between planes
The planes \(\Pi_{1}: x + y + z = 6\) and \(\Pi_{2}: 2x - y + z = 3\) meet in the line \(l\). Find a vector equation of \(l\) and the acute angle between the planes.
Find the acute angle between the planes \(x + 2y + 2z = 5\) and \(-2x + y - 2z = 1\).
\(\mathbf{n}_{1} \cdot \mathbf{n}_{2} = -2 + 2 - 4 = -4\), both normals have length 3, so \(\cos\theta = \dfrac{4}{9}\) and \(\theta = 63.6^\circ\). (Without the modulus you get the obtuse \(116.4^\circ\).)
Examiner trap
Using \(\cos\theta = \dfrac{|\mathbf{b} \cdot \mathbf{n}|}{|\mathbf{b}||\mathbf{n}|}\) for a line and a plane gives the angle with the NORMAL. In the first example that is \(63.6^\circ\), and the answer is \(90^\circ - 63.6^\circ = 26.4^\circ\).
Stage 2 · Exam
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Find the point of intersection of the line \(\mathbf{r} = \begin{pmatrix}2\\-1\\4\end{pmatrix} + t\begin{pmatrix}1\\3\\-2\end{pmatrix}\) and the plane \(2x + y + z = 1\).
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Find the foot of the perpendicular from \(A(3, -1, 4)\) to the plane \(x - y + 2z = 0\).
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Find the acute angle between the planes \(2x - y + 2z = 3\) and \(x + 2y - 2z = 1\).
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The line \(\mathbf{r} = \begin{pmatrix}1\\2\\0\end{pmatrix} + t\begin{pmatrix}2\\-1\\k\end{pmatrix}\) is parallel to the plane \(3x + 2y - z = 7\). Find \(k\) and describe the line.
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