Angles Between Lines and Planes

How steep is a roof? At what angle does a laser beam hit a mirror? Each of these questions asks for an angle between lines and planes in 3-D. You already know how to find the angle between two vectors with the scalar (dot) product. This page shows which two vectors to use in each case, and why the answer is always given as an acute angle.

This page writes vectors as columns, as IB does. The Ontario vector pages on this site write the same vectors in square brackets, so $\begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}$ there is $[2, -1, 2]$.

Key ideas

The tool: the angle between two vectors

For non-zero vectors $\vec{v}$ and $\vec{w}$, the angle $\theta$ between them satisfies

$$\cos\theta = \frac{\vec{v} \cdot \vec{w}}{\lvert\vec{v}\rvert\,\lvert\vec{w}\rvert}, \qquad 0^\circ \le \theta \le 180^\circ$$

Everything on this page uses this formula with the right pair of vectors: direction vectors for lines, and normal vectors for planes.

The angle between two lines

Two lines $\vec{r} = \vec{a}_1 + \lambda\vec{b}_1$ and $\vec{r} = \vec{a}_2 + \mu\vec{b}_2$ meet at the same angle as their direction vectors $\vec{b}_1$ and $\vec{b}_2$. But crossing lines make two angles, $\theta$ and $180^\circ - \theta$, and the angle between the lines is the acute one. Taking the absolute value of the scalar product gives it straight away:

$$\cos\theta = \frac{\lvert\vec{b}_1 \cdot \vec{b}_2\rvert}{\lvert\vec{b}_1\rvert\,\lvert\vec{b}_2\rvert}$$

If you forget the absolute value and get an obtuse angle, subtract it from $180^\circ$. The formula uses only directions, so it works for lines in 2-D or 3-D, and even for skew lines that never meet.

The position vectors $\vec{a}_1$ and $\vec{a}_2$ play no part. Don’t use them!

The angle between a line and a plane

The angle between a line and a plane is the angle $\theta$ between the line and its projection onto the plane (its “shadow” when light shines straight down onto the plane). It runs from $0^\circ$ (line parallel to the plane) to $90^\circ$ (line perpendicular to the plane).

A plane has no single direction vector, but it does have a normal $\vec{n}$. Let $\varphi$ be the acute angle between the line’s direction $\vec{b}$ and $\vec{n}$. In the figure, the normal is perpendicular to the projection, so

$$\theta = 90^\circ - \varphi$$

The angle $\theta$ between the line and the plane and the angle $\varphi$ between the line and the normal add to $90^\circ$.

Since $\cos\varphi = \sin(90^\circ - \varphi) = \sin\theta$, you can go straight to $\theta$ with sine:

$$\sin\theta = \frac{\lvert\vec{b} \cdot \vec{n}\rvert}{\lvert\vec{b}\rvert\,\lvert\vec{n}\rvert}$$

Two special cases are worth spotting before you calculate:

• $\vec{b} \cdot \vec{n} = 0$: the line is parallel to the plane (or lies in it), so $\theta = 0^\circ$.
• $\vec{b}$ is a multiple of $\vec{n}$: the line is perpendicular to the plane, so $\theta = 90^\circ$.

The angle between two planes

Two planes that meet do so along a line. Look along that line, so each plane appears as a line seen edge-on: each normal is perpendicular to its own plane, so the angle between the normals equals the angle between the planes. Again, use the acute one:

$$\cos\theta = \frac{\lvert\vec{n}_1 \cdot \vec{n}_2\rvert}{\lvert\vec{n}_1\rvert\,\lvert\vec{n}_2\rvert}$$

For a plane $ax + by + cz = d$, read the normal off the coefficients: $\vec{n} = \begin{pmatrix} a \\ b \\ c \end{pmatrix}$. For a plane in vector form $\vec{r} = \vec{a} + \lambda\vec{b} + \mu\vec{c}$, find the normal with the vector product: $\vec{n} = \vec{b} \times \vec{c}$.

Summary

Angle between	Vectors to use	Formula for the acute angle
two lines	directions $\vec{b}_1$, $\vec{b}_2$	$\cos\theta = \dfrac{\lvert\vec{b}_1 \cdot \vec{b}_2\rvert}{\lvert\vec{b}_1\rvert\lvert\vec{b}_2\rvert}$
a line and a plane	direction $\vec{b}$, normal $\vec{n}$	$\sin\theta = \dfrac{\lvert\vec{b} \cdot \vec{n}\rvert}{\lvert\vec{b}\rvert\lvert\vec{n}\rvert}$
two planes	normals $\vec{n}_1$, $\vec{n}_2$	$\cos\theta = \dfrac{\lvert\vec{n}_1 \cdot \vec{n}_2\rvert}{\lvert\vec{n}_1\rvert\lvert\vec{n}_2\rvert}$

Give angles in degrees to 3 significant figures unless the question asks for radians.

Worked examples

Example 1: Two lines in 3-D

Find the acute angle between the lines

$$L_1:\ \vec{r} = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} + \lambda\begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix} \qquad\text{and}\qquad L_2:\ \vec{r} = \begin{pmatrix} 0 \\ 3 \\ -1 \end{pmatrix} + \mu\begin{pmatrix} 1 \\ 2 \\ -2 \end{pmatrix}$$

Solution. Use the direction vectors only:

$$\vec{b}_1 \cdot \vec{b}_2 = 2(1) + (-1)(2) + 2(-2) = -4, \qquad \lvert\vec{b}_1\rvert = \sqrt{4 + 1 + 4} = 3, \qquad \lvert\vec{b}_2\rvert = \sqrt{1 + 4 + 4} = 3$$ $$\cos\theta = \frac{\lvert -4 \rvert}{3 \times 3} = \frac{4}{9} \quad\Rightarrow\quad \theta = \cos^{-1}\!\left(\frac{4}{9}\right) \approx 63.6^\circ \text{ (3 s.f.)}$$

Without the absolute value you’d get $\cos^{-1}\!\left(-\dfrac{4}{9}\right) \approx 116.4^\circ$, the obtuse angle between the lines. Check: $180^\circ - 116.4^\circ = 63.6^\circ$. ✓

Example 2: A line and a plane

Find the angle between the line $\vec{r} = \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix} + t\begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}$ and the plane $2x - y + z = 5$.

Solution. The direction of the line is $\vec{b} = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}$ and the normal to the plane is $\vec{n} = \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}$.

$$\vec{b} \cdot \vec{n} = 2 - 1 + 2 = 3, \qquad \lvert\vec{b}\rvert = \sqrt{6}, \qquad \lvert\vec{n}\rvert = \sqrt{6}$$ $$\sin\theta = \frac{\lvert 3 \rvert}{\sqrt{6}\sqrt{6}} = \frac{3}{6} = \frac{1}{2} \quad\Rightarrow\quad \theta = 30^\circ$$

Check with the normal. The angle between $\vec{b}$ and $\vec{n}$ satisfies $\cos\varphi = \dfrac{1}{2}$, so $\varphi = 60^\circ$, and $\theta = 90^\circ - 60^\circ = 30^\circ$. ✓

Example 3: Two planes

Find the acute angle between the planes $x + 2y - 2z = 4$ and $-3x + 4z = 1$.

Solution. The normals are $\vec{n}_1 = \begin{pmatrix} 1 \\ 2 \\ -2 \end{pmatrix}$ and $\vec{n}_2 = \begin{pmatrix} -3 \\ 0 \\ 4 \end{pmatrix}$. (The second plane has no $y$ term, so its $y$-component is $0$.)

$$\vec{n}_1 \cdot \vec{n}_2 = -3 + 0 - 8 = -11, \qquad \lvert\vec{n}_1\rvert = 3, \qquad \lvert\vec{n}_2\rvert = 5$$ $$\cos\theta = \frac{\lvert -11 \rvert}{3 \times 5} = \frac{11}{15} \quad\Rightarrow\quad \theta \approx 42.8^\circ \text{ (3 s.f.)}$$

The normals themselves are $137.2^\circ$ apart; the planes meet at $42.8^\circ$ and $137.2^\circ$, and the acute angle $42.8^\circ$ is the answer.

Example 4: Different forms of equations

Find the angle between the line $\dfrac{x - 1}{2} = \dfrac{y + 3}{-1} = \dfrac{z}{2}$ and the plane

$$\vec{r} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} + \lambda\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} + \mu\begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix}$$

Solution. Direction of the line. In Cartesian form, the direction vector is made of the denominators: $\vec{b} = \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}$.

Normal to the plane. The two direction vectors lie in the plane, so their vector product is normal to it:

$$\vec{n} = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} \times \begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix} = \begin{pmatrix} (1)(-1) - (0)(1) \\ (0)(0) - (1)(-1) \\ (1)(1) - (1)(0) \end{pmatrix} = \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}$$

Check: $\vec{n} \cdot \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} = -1 + 1 + 0 = 0$ and $\vec{n} \cdot \begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix} = 0 + 1 - 1 = 0$. ✓

The angle.

$$\vec{b} \cdot \vec{n} = -2 - 1 + 2 = -1, \qquad \lvert\vec{b}\rvert = 3, \qquad \lvert\vec{n}\rvert = \sqrt{3}$$ $$\sin\theta = \frac{\lvert -1 \rvert}{3\sqrt{3}} \approx 0.19245 \quad\Rightarrow\quad \theta \approx 11.1^\circ \text{ (3 s.f.)}$$

The line is close to parallel to the plane, which matches the small value of $\vec{b} \cdot \vec{n}$.

Common mistakes

Using cosine for the angle between a line and a plane. $\dfrac{\lvert\vec{b} \cdot \vec{n}\rvert}{\lvert\vec{b}\rvert\lvert\vec{n}\rvert}$ is the cosine of the angle with the normal. Either take $\sin^{-1}$ of it, or take $\cos^{-1}$ and subtract from $90^\circ$. In Example 2, $\cos^{-1}\!\left(\dfrac{1}{2}\right) = 60^\circ$ is the angle with the normal, not the answer.

Giving the obtuse angle. The angle between two lines, or two planes, is the acute one. Use the absolute value of the scalar product, or subtract an obtuse answer from $180^\circ$.

Using position vectors instead of direction vectors. The point a line passes through has nothing to do with its direction. Only $\vec{b}_1$ and $\vec{b}_2$ go into the formula, never $\vec{a}_1$ or $\vec{a}_2$.

Reading the wrong numbers from a Cartesian equation. For the line $\dfrac{x - 1}{2} = \dfrac{y + 3}{-1} = \dfrac{z}{2}$, the direction is the denominators $\begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}$, while $(1, -3, 0)$ is a point. For the plane $x + 2y - 2z = 4$, the normal is the coefficients $\begin{pmatrix} 1 \\ 2 \\ -2 \end{pmatrix}$, and a missing variable means a component of $0$.

Using a direction vector of a plane as if it were the normal. For a plane in the form $\vec{r} = \vec{a} + \lambda\vec{b} + \mu\vec{c}$, the vectors $\vec{b}$ and $\vec{c}$ lie in the plane. Find $\vec{n} = \vec{b} \times \vec{c}$ first.

Calculator in radians. If $\cos^{-1}\!\left(\dfrac{4}{9}\right)$ comes out as $1.11$, your calculator is in radian mode. That’s correct in radians, but give degrees unless the question asks otherwise.

Practice

1. (Warm-up) Find the acute angle between the lines $\vec{r} = \begin{pmatrix} 0 \\ 1 \end{pmatrix} + \lambda\begin{pmatrix} 3 \\ 4 \end{pmatrix}$ and $\vec{r} = \begin{pmatrix} 2 \\ 0 \end{pmatrix} + \mu\begin{pmatrix} 5 \\ -12 \end{pmatrix}$.

Solution

$$\vec{b}_1 \cdot \vec{b}_2 = 15 - 48 = -33, \qquad \lvert\vec{b}_1\rvert = 5, \qquad \lvert\vec{b}_2\rvert = 13$$$$\cos\theta = \frac{33}{65} \quad\Rightarrow\quad \theta \approx 59.5^\circ \text{ (3 s.f.)}$$

2. (Warm-up) Without finding any inverse cosines, decide whether each line is parallel or perpendicular to the plane $x - 2y + 3z = 1$, and state the angle between them.

• (a) A line with direction $\begin{pmatrix} 2 \\ -4 \\ 6 \end{pmatrix}$.
• (b) A line with direction $\begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}$.

Solution

The normal is $\vec{n} = \begin{pmatrix} 1 \\ -2 \\ 3 \end{pmatrix}$.

(a) $\begin{pmatrix} 2 \\ -4 \\ 6 \end{pmatrix} = 2\vec{n}$, so the line is parallel to the normal, which means it is perpendicular to the plane: the angle is $90^\circ$.

(b) $\begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} \cdot \vec{n} = 1 - 4 + 3 = 0$, so the line is perpendicular to the normal, which means it is parallel to the plane (or lies in it): the angle is $0^\circ$.

3. (Core) Line $L_1$ passes through $A(1, 2, 3)$ and $B(3, 3, 5)$. Line $L_2$ passes through $C(0, 1, -1)$ and $D(4, -1, 3)$. Find the acute angle between the lines.

Solution

Direction vectors: $\overrightarrow{AB} = \begin{pmatrix} 2 \\ 1 \\ 2 \end{pmatrix}$ and $\overrightarrow{CD} = \begin{pmatrix} 4 \\ -2 \\ 4 \end{pmatrix}$, which is $2\begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}$. A multiple doesn’t change the direction, so use $\begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}$:

$$\cos\theta = \frac{\lvert 4 - 1 + 4 \rvert}{3 \times 3} = \frac{7}{9} \quad\Rightarrow\quad \theta \approx 38.9^\circ \text{ (3 s.f.)}$$

4. (Core) A ski lift cable follows the line $\vec{r} = \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix} + t\begin{pmatrix} 3 \\ 0 \\ 4 \end{pmatrix}$, where $z$ is the height. Find the angle the cable makes with the horizontal plane $z = 0$.

Solution

The plane $z = 0$ has normal $\vec{n} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$, and $\vec{b} = \begin{pmatrix} 3 \\ 0 \\ 4 \end{pmatrix}$ has magnitude $5$.

$$\sin\theta = \frac{\lvert 4 \rvert}{5 \times 1} = 0.8 \quad\Rightarrow\quad \theta \approx 53.1^\circ \text{ (3 s.f.)}$$

Check: the cable rises $4$ for every $3$ across, and $\tan^{-1}\!\left(\dfrac{4}{3}\right) \approx 53.1^\circ$. ✓

5. (Core) Find the acute angle between the planes $2x + y - z = 3$ and $x - y + 2z = 0$.

Solution

$\vec{n}_1 = \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}$ and $\vec{n}_2 = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}$, each with magnitude $\sqrt{6}$.

$$\vec{n}_1 \cdot \vec{n}_2 = 2 - 1 - 2 = -1, \qquad \cos\theta = \frac{\lvert -1 \rvert}{\sqrt{6}\sqrt{6}} = \frac{1}{6} \quad\Rightarrow\quad \theta \approx 80.4^\circ \text{ (3 s.f.)}$$

6. (Core) A flat roof panel contains the points $(0, 0, 3)$, $(8, 0, 3)$ and $(0, 4, 6)$, in metres, where $z$ is the height above the floor $z = 0$. Find the angle between the roof panel and the floor.

Solution

Two vectors in the roof: $\begin{pmatrix} 8 \\ 0 \\ 0 \end{pmatrix}$ and $\begin{pmatrix} 0 \\ 4 \\ 3 \end{pmatrix}$. A normal is

$$\begin{pmatrix} 8 \\ 0 \\ 0 \end{pmatrix} \times \begin{pmatrix} 0 \\ 4 \\ 3 \end{pmatrix} = \begin{pmatrix} 0 \cdot 3 - 0 \cdot 4 \\ 0 \cdot 0 - 8 \cdot 3 \\ 8 \cdot 4 - 0 \cdot 0 \end{pmatrix} = \begin{pmatrix} 0 \\ -24 \\ 32 \end{pmatrix} = 8\begin{pmatrix} 0 \\ -3 \\ 4 \end{pmatrix}$$

Use $\vec{n}_1 = \begin{pmatrix} 0 \\ -3 \\ 4 \end{pmatrix}$ (magnitude $5$) and the floor’s normal $\vec{n}_2 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$:

$$\cos\theta = \frac{\lvert 4 \rvert}{5 \times 1} = 0.8 \quad\Rightarrow\quad \theta \approx 36.9^\circ \text{ (3 s.f.)}$$

Check: the roof rises $3$ m over a horizontal run of $4$ m, and $\tan^{-1}\!\left(\dfrac{3}{4}\right) \approx 36.9^\circ$. ✓

7. (Core) A line has direction $\begin{pmatrix} 1 \\ k \\ 2 \end{pmatrix}$, and $\Pi$ is the plane $3x - y + z = 4$.

• (a) Find the value of $k$ for which the line is parallel to $\Pi$.
• (b) Show that there is no value of $k$ for which the line is perpendicular to $\Pi$.

Solution

The normal is $\vec{n} = \begin{pmatrix} 3 \\ -1 \\ 1 \end{pmatrix}$.

(a) Parallel to the plane means perpendicular to the normal: $3 - k + 2 = 0$, so $k = 5$.

(b) Perpendicular to the plane means $\begin{pmatrix} 1 \\ k \\ 2 \end{pmatrix} = c\begin{pmatrix} 3 \\ -1 \\ 1 \end{pmatrix}$ for some $c$. The first component gives $c = \dfrac{1}{3}$, but the third gives $c = 2$. These contradict each other, so no value of $k$ works.

8. (Challenge) The line through the origin with direction $\begin{pmatrix} 1 \\ 1 \\ p \end{pmatrix}$ makes an angle of $45^\circ$ with the plane $z = 0$. Find the possible values of $p$.

Solution

With $\vec{n} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$:

$$\sin 45^\circ = \frac{\lvert p \rvert}{\sqrt{2 + p^2}}$$

Square both sides, using $\sin^2 45^\circ = \dfrac{1}{2}$:

$$\begin{aligned} \frac{1}{2} &= \frac{p^2}{2 + p^2} \\ 2 + p^2 &= 2p^2 \\ p^2 &= 2 \end{aligned}$$

So $p = \sqrt{2}$ or $p = -\sqrt{2}$. (One line points upward and one downward; both make $45^\circ$ with the plane.)

Check: $\dfrac{\sqrt{2}}{\sqrt{4}} = \dfrac{\sqrt{2}}{2} = \sin 45^\circ$. ✓

9. (Challenge) The planes $x + y + z = 1$ and $x - y = 0$ meet in a line $L$. Find the angle between $L$ and the plane $2x + y + z = 7$.

Solution

$L$ lies in both planes, so it is perpendicular to both normals. Its direction is their vector product:

$$\vec{b} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \times \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix} = \begin{pmatrix} (1)(0) - (1)(-1) \\ (1)(1) - (1)(0) \\ (1)(-1) - (1)(1) \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ -2 \end{pmatrix}$$

With $\vec{n} = \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}$:

$$\vec{b} \cdot \vec{n} = 2 + 1 - 2 = 1, \qquad \lvert\vec{b}\rvert = \lvert\vec{n}\rvert = \sqrt{6}$$$$\sin\theta = \frac{1}{6} \quad\Rightarrow\quad \theta \approx 9.59^\circ \text{ (3 s.f.)}$$