Distances, angles and volumes in space

Summary

This lesson brings together the tools of the three preceding ones, the dot product, the cross product and the scalar triple product, with the equations of lines and planes, in order to measure in space. It defines the distance from a point to a set and proves, by projecting onto the normal, the formula for the distance from a point to a plane, and, with the cross product, the formula for the distance from a point to a line, both with the foot of the perpendicular as the only closest point. From them it deduces the distances between parallel planes and between parallel lines, and it proves that two skew lines have a unique common perpendicular, which is found by solving a system of two equations and whose length is a scalar triple product divided by a norm. It defines and computes the angles between two lines, between a line and a plane and between two planes, and proves that they coincide with the angles of elementary geometry; it constructs the perpendicular bisector plane of a line segment and the reflections of a point in a plane and in a line; it proves that the area of the projection of a plane polygon is its area multiplied by the cosine of the dihedral angle, and it studies the sphere, its equation, its intersection with a plane and its tangent plane. The applications are the planes of a crystal with their Miller indices and satellite positioning; the problems deal with the regular tetrahedron, the points equidistant from three points, the shadow of a solar panel, the tangent planes to a sphere that contain a line and two aircraft whose routes are skew.

Learning objectives

Compute the distance from a point to a plane and to a line, between parallel planes, between parallel lines and between skew lines, proving each formula and finding the points at which it is attained.
Find the common perpendicular to two skew lines and prove that it is unique.
Compute angles between lines, between a line and a plane and between two planes, and prove that they coincide with the geometric angles they define.
Construct the perpendicular bisector plane of a line segment and the reflections of a point in a plane and in a line.
Compute areas and volumes with the cross and scalar triple products, and prove and apply the relation between the area of a plane polygon and that of its projection.
Obtain the center and the radius of a sphere from its equation, decide its position relative to a plane, find the circle of intersection and the tangent plane, and model with these tools problems in crystallography, solar energy, air navigation and satellite positioning.

The distance from a point to a plane and to a line

The lesson on vectors in the plane obtained the distance from a point to a line in the plane by projecting onto the normal vector. In space, the lesson on lines and planes in space wrote the plane through $P_0$ with a nonzero normal vector $\vec{n} = \left( a, b, c \right)$ as the set of the points $X$ with $\vec{n} \cdot \vec{P_0 X} = 0$ , that is, with an equation $ax + by + cz = d$ , and the line through $A$ with direction vector $\vec{u}$ as the set of the points $A + t \vec{u}$ , with $t$ real. If this is so, the idea from the plane carries over unchanged to planes in space; for lines, however, the normal is no longer a single direction, and the cross product from the lesson on the cross product and the scalar triple product takes its place. The facts of Euclidean solid geometry that are used are those admitted in the lesson on vectors in space and the dot product.

Let $F$ be a nonempty set of points in space and $P$ a point. If among the distances from $P$ to the points of $F$ there is one that is less than or equal to all the others, that number is the distance from $P$ to $F$ , and it is written $\operatorname{d}(P, F)$ . Likewise, if among the distances from the points of a set $F$ to those of another set $G$ there is one less than or equal to all of them, that number is the distance between $F$ and $G$ , $\operatorname{d}(F, G)$ .

The definition does not guarantee that the distance exists: among infinitely many distances there may be no smallest one. That is why the theorems that follow do not merely give a formula; they prove that there is a closest point, and that it is unique.

TheoremDistance from a point to a plane

Let $\pi$ be the plane $ax + by + cz = d$ , with $\vec{n} = \left( a, b, c \right)$ nonzero, and let $P = (x_0, y_0, z_0)$ . The point $H = P - \lambda \vec{n}$ , with $\lambda = \frac{a x_0 + b y_0 + c z_0 - d}{{\norma{\vec{n}}}^2}$ , lies in $\pi$ , $\vec{HP}$ is normal to $\pi$ , and $H$ is the only point of $\pi$ closest to $P$ . Consequently:

\operatorname{d}(P, \pi) = \frac{\abs{a x_0 + b y_0 + c z_0 - d}}{\sqrt{a^2 + b^2 + c^2}}

The vector from an arbitrary point of the plane to $P$ is projected onto the normal, as in the plane; minimality is the Pythagorean theorem. The figure uses the plane $2x + y + 2z = 4$ and the point $P = (2, 3, 3)$ .

Proof

$\vec{p} = \dato{p}{\frac{\vec{QP} \cdot \vec{n}}{{\norma{\vec{n}}}^2} \vec{n}}$
We choose a point $Q = (q_1, q_2, q_3)$ of $\pi$ and project $\vec{QP}$ onto $\vec{n}$ with the orthogonal projection theorem from the lesson on vectors in space. In the figure, $Q = (2, 0, 0)$ .
$\vec{QP} \cdot \vec{n} = a x_0 + b y_0 + c z_0 \resaltar{- \left( a q_1 + b q_2 + c q_3 \right)} = \dato{e}{a x_0 + b y_0 + c z_0 - \resaltar{d}}$
Since $Q$ lies in $\pi$ , $a q_1 + b q_2 + c q_3 = d$ , and the dot product in the numerator is the left-hand side of the equation evaluated at $P$ , minus $d$ .
$\vec{QH} = \vec{QP} - \resaltar{\lambda \vec{n}}, \qquad \vec{n} \cdot \vec{QH} = 0 \rightarrow \dato{h}{H \in \pi}$
Therefore, $\vec{p} = \lambda \vec{n}$ and $H = P - \vec{p}$ . The vector $\vec{QH} = \vec{QP} - \vec{p}$ is the orthogonal component of the projection, perpendicular to $\vec{n}$ ; by the point-normal equation of the plane, $H$ lies in $\pi$ .
${\norma{\vec{PX}}}^2 = {\norma{\vec{PH}}}^2 + \resaltar{{\norma{\vec{HX}}}^2} \ge {\norma{\vec{PH}}}^2$
Let $X$ be any point of $\pi$ . By the Chasles relation, $\vec{PX} = \vec{PH} + \vec{HX}$ ; the first summand is a multiple of $\vec{n}$ and the second joins two points of $\pi$ , so they are orthogonal, and the Pythagorean equality holds.
$\operatorname{d}(P, \pi) = \norma{\vec{p}} = \frac{\resaltar{\abs{\vec{QP} \cdot \vec{n}}}}{\norma{\vec{n}}} = \frac{\abs{a x_0 + b y_0 + c z_0 - d}}{\sqrt{a^2 + b^2 + c^2}}$
Equality holds only if $\vec{HX} = \vec{0}$ , that is, if $X = H$ : the foot of the perpendicular is the only closest point, and the distance is the norm of the projection.
$\operatorname{d}(P, \pi) = \frac{\abs{2 \cdot 2 + 3 + 2 \cdot 3 - 4}}{\sqrt{4 + 1 + 4}} = \frac{\resaltar{9}}{\resaltar{3}} = 3$
In the figure, $\vec{QP} = \left( 0, 3, 3 \right)$ , $\vec{QP} \cdot \vec{n} = 9$ and $\norma{\vec{n}} = 3$ , so that $\lambda = 1$ , the foot is $H = (0, 2, 1)$ and the distance is $3$ .

The sign of the number $a x_0 + b y_0 + c z_0 - d$ , which the formula hides with the absolute value, also has a meaning. Since $P = H + \lambda \vec{n}$ , it is positive when $P$ is reached from the plane by moving in the direction of $\vec{n}$ , and negative when it is reached in the opposite direction. The points with $ax + by + cz > d$ form one of the two half-spaces that the plane determines, and the points with $ax + by + cz < d$ form the other; two points are on the same side of the plane if and only if the expression has the same sign at both. Thus, the origin and the point $(2, 3, 3)$ lie on opposite sides of the plane $2x + y + 2z = 4$ , because $0 - 4 < 0 < 9$ .

TheoremDistance from a point to a line

Let $r$ be the line $A + t \vec{u}$ , with $\vec{u}$ nonzero, and let $P$ be a point. The point $H = A + t_0 \vec{u}$ , with $t_0 = \frac{\vec{AP} \cdot \vec{u}}{{\norma{\vec{u}}}^2}$ , is the only point of $r$ closest to $P$ , $\vec{HP}$ is perpendicular to $\vec{u}$ , and:

\operatorname{d}(P, r) = \frac{\norma{\vec{AP} \times \vec{u}}}{\norma{\vec{u}}}

One decomposes $\vec{AP}$ into its projection onto $\vec{u}$ and its orthogonal component; the norm of the latter is expressed with Lagrange’s identity. The figure uses the line through $A = (0, 0, 1)$ with direction vector $\vec{u} = \left( 1, 2, 2 \right)$ and the point $P = (3, 3, 1)$ .

Proof

$\vec{AP} = t_0 \vec{u} + \dato{w}{\vec{w}}, \qquad \vec{w} \cdot \vec{u} = 0$
By the orthogonal projection theorem, $\vec{AP}$ is the sum of its projection $t_0 \vec{u}$ onto $\vec{u}$ and of a vector $\vec{w}$ orthogonal to $\vec{u}$ . The point $H = A + t_0 \vec{u}$ lies on $r$ and $\vec{HP} = \vec{w}$ .
$\vec{XP} = \vec{w} + \resaltar{\left( t_0 - t \right) \vec{u}}, \qquad {\norma{\vec{XP}}}^2 = {\norma{\vec{w}}}^2 + \resaltar{\left( t_0 - t \right)^2 {\norma{\vec{u}}}^2} \ge {\norma{\vec{w}}}^2$
For an arbitrary point $X = A + t \vec{u}$ of $r$ , the vector $\vec{XP}$ is $\vec{w}$ plus a multiple of $\vec{u}$ ; the two summands are orthogonal and the Pythagorean equality holds.
${\operatorname{d}(P, r)}^2 = {\norma{\vec{w}}}^2 = {\norma{\vec{AP}}}^2 - \resaltar{\frac{\left( \vec{AP} \cdot \vec{u} \right)^2}{{\norma{\vec{u}}}^2}} = \dato{q}{\frac{{\norma{\vec{AP}}}^2 {\norma{\vec{u}}}^2 - \left( \vec{AP} \cdot \vec{u} \right)^2}{{\norma{\vec{u}}}^2}}$
Equality holds only if $t = t_0$ : $H$ is the only closest point and the distance is the norm of $\vec{w}$ , whose square is given by the Pythagorean equality of the projection.
${\operatorname{d}(P, r)}^2 = \frac{\resaltar{{\norma{\vec{AP} \times \vec{u}}}^2}}{{\norma{\vec{u}}}^2} \rightarrow \operatorname{d}(P, r) = \frac{\norma{\vec{AP} \times \vec{u}}}{\norma{\vec{u}}}$
By Lagrange’s identity from the lesson on the cross product, the numerator is the square of the norm of $\vec{AP} \times \vec{u}$ ; we take the square root.
$\vec{AP} \times \vec{u} = \left( 6, -6, 3 \right), \qquad \operatorname{d}(P, r) = \frac{\resaltar{9}}{\resaltar{3}} = 3$
In the figure, $\vec{AP} = \left( 3, 3, 0 \right)$ and $\vec{AP} \cdot \vec{u} = 9$ , so that $t_0 = 1$ , $H = (1, 2, 3)$ and $\vec{w} = \left( 2, 1, -2 \right)$ , of norm $3$ . The cross product gives the same.

The formula has a direct geometric reading: $\norma{\vec{AP} \times \vec{u}}$ is the area of the parallelogram with sides $\vec{AP}$ and $\vec{u}$ , and $\norma{\vec{u}}$ is the length of its base on the line; the quotient is the height, which is the required distance. When only the distance matters, the cross product avoids computing the foot of the perpendicular.

In the rest of this lesson

Distances between parallel planes and between parallel lines
The distance between two skew lines
Angles between lines and planes
The perpendicular bisector plane and reflections
Areas and volumes
The sphere
Applications
The separation between two pipelines
The planes of a crystal and Miller indices
Satellite positioning
Solved problems
The opposite edges of a regular tetrahedron
The points equidistant from three points
The shadow of a solar panel
The tangent planes that contain a line
Two aircraft whose routes are skew
Proposed problems
The hip roof
The bisector planes
A ray of light and a plane mirror
The sphere inscribed in a tetrahedron
The volume from two opposite edges