Angle Between Planes / Line and Plane, and Distance of a Point from a Plane

MATAngles & Distances

Angles in 3D between flat objects are computed from their normals and directions: the angle between two planes uses their normals, while the angle between a line and a plane uses the complement (line direction versus normal). The perpendicular distance of a point from a plane, the foot of the perpendicular, and the image (reflection) of a point are all standard ISC computations built on these ideas.

Angle between two planes

\cos\theta = \dfrac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}|\,|\vec{n_2}|}

\vec{n_1}, \vec{n_2}

are the normal vectors of the two planes; the modulus gives the acute angle between them.

Angle between a line and a plane

\sin\theta = \dfrac{|\vec{b} \cdot \vec{n}|}{|\vec{b}|\,|\vec{n}|}

\vec{b}

is the line's direction,

\vec{n}

the plane's normal;

\theta

is the angle between the line and the plane (NOT the normal).

Distance of a point from a plane

d = \dfrac{|a x_1 + b y_1 + c z_1 + d_0|}{\sqrt{a^2 + b^2 + c^2}}

Plane

ax+by+cz+d_0=0

and point

P(x_1,y_1,z_1)

; the modulus ensures a non-negative distance.

Distance between two parallel planes

d = \dfrac{|d_1 - d_2|}{\sqrt{a^2 + b^2 + c^2}}

Parallel planes

ax+by+cz+d_1=0

and

ax+by+cz+d_2=0

written with identical normal coefficients.

Foot of perpendicular and image of a point

\dfrac{x - x_1}{a} = \dfrac{y - y_1}{b} = \dfrac{z - z_1}{c} = -\dfrac{a x_1 + b y_1 + c z_1 + d_0}{a^2 + b^2 + c^2}

(x,y,z)

is the foot of the perpendicular from

P(x_1,y_1,z_1)

to plane

ax+by+cz+d_0=0

; for the image, replace the right-hand fraction by

-2

times the same quantity.

Two planes are parallel when their normals are proportional, and perpendicular when $\vec{n_1}\cdot\vec{n_2}=0$ .
For a line and plane, $\sin\theta$ (not $\cos\theta$ ) is used because $\theta$ is the complement of the angle between the line and the normal.
A line is parallel to a plane when $\vec{b}\cdot\vec{n}=0$ , and perpendicular to the plane when $\vec{b}$ is parallel to $\vec{n}$ .
Before using the parallel-plane distance formula, rewrite both planes with the SAME normal coefficients (scale one equation if needed).
Foot-of-perpendicular method: write the line through $P$ along the normal as $x=x_1+a\lambda$ etc., substitute into the plane to solve for $\lambda$ , then read off the foot.
Image (mirror reflection) $P'$ satisfies: the foot $M$ is the midpoint of $PP'$ , so $P' = 2M - P$ componentwise.
The perpendicular distance equals $|\lambda|\sqrt{a^2+b^2+c^2}$ , where $\lambda$ is the parameter value at the foot — a useful cross-check.
Always simplify the surd and state distances in the given units (e.g. metres).

Where the marks go

Using $\cos\theta$ instead of $\sin\theta$ for the angle between a line and a plane (or vice versa) — recall $\theta_{\text{line-plane}} = 90^\circ - \theta_{\text{line-normal}}$ .
Dropping the modulus in the distance formula, producing a negative 'distance'.
Forgetting to make the two parallel planes' normal coefficients identical before applying the gap formula, giving a wrong $|d_1-d_2|$ .
Computing the image by stopping at the foot of the perpendicular, or using $-1$ times instead of $-2$ times the fraction, so $P'$ is wrong.

How the board asks it

Numericaldistance of a point from a plane
Find the perpendicular distance of the point $(2,5,-3)$ from the plane $6x-3y+2z-5=0$ , and the distance between the parallel planes $2x-2y+z+3=0$ and $4x-4y+2z+5=0$ .
Numericalangle between two planes
Find the angle between the planes $2x-y+2z=5$ and $3x+6y-2z=7$ , and hence state whether the two planes are perpendicular.
Numericalangle between a line and a plane
Find the angle between the line $\dfrac{x-1}{2}=\dfrac{y+1}{-1}=\dfrac{z}{2}$ and the plane $x+2y+2z-5=0$ , using $\sin\theta$ for the line-plane angle.
Numericalfoot of perpendicular and image of a point
Find the coordinates of the foot of the perpendicular and the image (reflection) of the point $P(1,3,4)$ in the plane $2x-y+z+3=0$ .
Give reasonsparallel and perpendicular conditions
Show that the line $\vec{r}=(\hat{i}+2\hat{j}+3\hat{k})+\lambda(2\hat{i}+3\hat{j}+4\hat{k})$ is parallel to the plane $\vec{r}\cdot(\hat{i}+2\hat{j}-2\hat{k})=5$ , and find the distance between the line and the plane.

Practise this topic

PreviousEquation of a Plane

Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.