Angle between line and plane; line in plane

When a line meets a plane, the angle between them is the complement of the angle between the line's direction and the plane's normal. This insight turns the line-plane angle into a dot-product calculation.

The angle

Line with direction $\vec d$; plane with normal $\vec n$. Let $\phi$ be the angle between $\vec d$ and $\vec n$, so $\cos\phi = \dfrac{\vec d\cdot\vec n}{|\vec d||\vec n|}$. The angle between the line and the plane is $\theta = \pi/2 - \phi$ (the line tilts away from the plane perpendicular).

So $\sin\theta = \cos\phi = \frac{|\vec d \cdot \vec n|}{|\vec d|\,|\vec n|}.$

The absolute value ensures $\theta \in [0, \pi/2]$.

Line parallel to plane

The line is parallel to the plane iff $\theta = 0$, i.e. $\vec d \cdot \vec n = 0$ , the direction is perpendicular to the normal. Geometrically, the line either lies in the plane or never meets it.

Line in plane

A line lies in the plane iff:

1. Its direction is perpendicular to the plane's normal: $\vec d\cdot\vec n = 0$.
2. Any one point on the line lies on the plane.

(The first condition alone gives a line parallel to the plane; the second pins it down.)

Line perpendicular to plane

The line is perpendicular to the plane iff its direction is parallel to the plane's normal: $\vec d \parallel \vec n$, i.e. $\vec d\times\vec n = \vec 0$. Then $\theta = \pi/2$.

Worked examples

Example 1. Find the angle between $\dfrac{x - 1}{2} = \dfrac{y}{3} = \dfrac{z + 2}{4}$ and the plane $2x - 3y + z = 5$.

$\vec d = (2, 3, 4)$, $\vec n = (2, -3, 1)$. $\vec d\cdot\vec n = 4 - 9 + 4 = -1$. $|\vec d| = \sqrt{29}$, $|\vec n| = \sqrt{14}$. $\sin\theta = \dfrac{1}{\sqrt{29}\cdot\sqrt{14}} = \dfrac{1}{\sqrt{406}}$.

Example 2. Show that $\dfrac{x - 1}{1} = \dfrac{y + 1}{2} = \dfrac{z - 1}{-1}$ is parallel to the plane $x + 2y - z = 5$.

$\vec d\cdot\vec n = 1 + 4 + 1 = 6 \ne 0$. Not parallel. (Direct check shows the angle is not zero.) Let me redo: $\vec d = (1, 2, -1)$, $\vec n = (1, 2, -1)$. They're identical , so the direction is the normal, meaning the line is perpendicular to the plane, not parallel. Different example: take line $\dfrac{x}{1} = \dfrac{y}{1} = \dfrac{z}{0}$ (direction $(1, 1, 0)$) and plane $z = 5$ (normal $(0, 0, 1)$). $\vec d\cdot\vec n = 0$. Parallel.

Example 3. Find the foot of perpendicular from $(1, 2, 3)$ to the plane $x + y + z = 6$.

The perpendicular from the point has direction = normal direction $(1, 1, 1)$. Parametrise: $\vec r = (1, 2, 3) + t(1, 1, 1) = (1 + t, 2 + t, 3 + t)$. On the plane: $(1 + t) + (2 + t) + (3 + t) = 6 + 3t = 6$, so $t = 0$.

Wait , that puts the foot at the original point. Let me recheck: $1 + 2 + 3 = 6$, so the point is on the plane already! Foot of perpendicular is the point itself.

Try $(1, 2, 0)$ instead: $1 + 2 + 0 = 3 \ne 6$. Foot: $(1 + t, 2 + t, t)$, on plane: $1 + t + 2 + t + t = 3 + 3t = 6$, $t = 1$. Foot: $(2, 3, 1)$.

Example 4. Find the image of $(1, -2, 1)$ in the plane $x - 2y + 2z + 3 = 0$.

Normal $(1, -2, 2)$, magnitude $3$. Foot of perpendicular: $(1, -2, 1) + t(1, -2, 2)$ on the plane: $(1 + t) - 2(-2 - 2t) + 2(1 + 2t) + 3 = 1 + t + 4 + 4t + 2 + 4t + 3 = 10 + 9t = 0$, so $t = -10/9$.

Foot: $(1 - 10/9, -2 + 20/9, 1 - 20/9) = (-1/9, 2/9, -11/9)$. Image: $(1, -2, 1) + 2 \cdot t \cdot \vec n / |\vec n|^2$ no wait , image is twice as far as foot from original. So image $= 2\cdot\text{foot} - (1, -2, 1) = (-2/9 - 1, 4/9 + 2, -22/9 - 1) = (-11/9, 22/9, -31/9)$.

Example 5. Show that the line $\dfrac{x - 1}{1} = \dfrac{y + 1}{2} = \dfrac{z - 2}{1}$ lies in the plane $2x - y - z = 1$.

Check perpendicularity of direction to normal: $\vec d\cdot\vec n = 2 - 2 - 1 = -1 \ne 0$. Not in the plane. (If we had asked correctly, we'd find the line meets the plane at a single point.)

Try line with direction $(1, 1, 1)$ and plane $x - y = 0$. Direction dotted with normal $(1, -1, 0)$: $1 - 1 + 0 = 0$. Take point on line: $(0, 0, 0)$. On plane? $0 - 0 = 0$, yes. So this line does lie in the plane.

Example 6. Find $\lambda$ if the line $\dfrac{x}{2} = \dfrac{y - 1}{\lambda} = \dfrac{z + 1}{3}$ is parallel to the plane $3x - y + 2z = 6$.

$\vec d\cdot\vec n = 6 - \lambda + 6 = 12 - \lambda = 0$, so $\lambda = 12$.

Try it yourself

1. Find the angle between $\dfrac{x}{1} = \dfrac{y}{1} = \dfrac{z}{1}$ and the plane $x + 2y + 3z = 6$.
2. Show that the $z$-axis is perpendicular to the plane $z = 0$.
3. Find $\lambda$ if $\dfrac{x}{1} = \dfrac{y - 1}{\lambda} = \dfrac{z - 2}{1}$ is parallel to $x + 2y + 3z = 6$.
4. Find the foot of perpendicular from $(2, 3, 4)$ to $x + 2y + 3z = 10$.
5. Image of $(0, 0, 0)$ in the plane $x + y + z = 3$.
6. Find the plane perpendicular to the line $\vec r = (1, 1, 1) + t(2, 1, 0)$ passing through $(0, 0, 1)$.
7. Show that the line $\vec r = (1, 2, 3) + t(2, 4, 6)$ is parallel to the plane $x + y - z = 5$ , or determine the relationship.
8. A line through $(1, 1, 1)$ is perpendicular to $3x + 4y - z = 12$. Find its direction.
9. Find the equation of the line through $(1, 0, 0)$ perpendicular to $2x + y - z = 0$.
10. Find $\mu$ such that the line $\vec r = (1, 2, 3) + t(1, \mu, 2)$ lies in the plane $x + 2y + z = 8$.
11. Find the perpendicular distance from $(2, 3, -4)$ to the plane $2x - 6y + 3z = 0$.
12. Find the angle between the line through $(1, 2, 3), (4, 5, 6)$ and the plane $x + y + z = 1$.
13. The line $\dfrac{x - 1}{1} = \dfrac{y + 2}{-1} = \dfrac{z}{2}$ meets the plane $x + y + z = 6$ at what point?
14. Show that the line $\vec r = (1, 2, -3) + t(2, 3, -1)$ lies entirely in the plane $x - y + z = -4$ , verify by checking both conditions.

Pitfalls and tricks

• Use $\sin$ for line-plane angle, $\cos$ for plane-plane and line-line angles. Don't confuse them.
• Parallel to plane $\Leftrightarrow$ $\vec d \cdot \vec n = 0$.
• In the plane $\Leftrightarrow$ parallel AND any point on the line is on the plane.
• Perpendicular to plane $\Leftrightarrow$ direction parallel to the normal.
• For foot of perpendicular from a point to a plane: parametrise along the normal, intersect with plane.