Distance from a point to a plane; intersections

The two remaining building blocks: distance from a point to a plane, and finding where two objects meet , line-plane intersection, two-plane intersection, etc.

Distance from a point to a plane

Vector form. Plane: $\vec r\cdot\vec n = d$. Point: $\vec p$. The distance is $D = \frac{|\vec p\cdot\vec n - d|}{|\vec n|}.$

Cartesian form. Plane: $Ax + By + Cz + D = 0$. Point: $(x_0, y_0, z_0)$. Distance: $D = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}.$

Why? The foot of perpendicular from $\vec p$ to the plane is $\vec p - \dfrac{\vec p\cdot\vec n - d}{|\vec n|^2}\vec n$. The distance equals the magnitude of that projection.

Line meets plane

Line: $\vec r = \vec a + t\vec d$. Plane: $\vec r\cdot\vec n = e$.

Substitute: $(\vec a + t\vec d)\cdot\vec n = e \Rightarrow t = \dfrac{e - \vec a\cdot\vec n}{\vec d\cdot\vec n}$.

If $\vec d\cdot\vec n \ne 0$, this gives the unique $t$ at which the line meets the plane. Plug $t$ back into the line's equation to find the point.

If $\vec d\cdot\vec n = 0$, the line is parallel to the plane. Then:

• $\vec a\cdot\vec n = e$: line lies in the plane.
• $\vec a\cdot\vec n \ne e$: line never meets the plane.

Two planes meet in a line

Planes $\vec r\cdot\vec n_1 = e_1$ and $\vec r\cdot\vec n_2 = e_2$ with $\vec n_1 \not\parallel \vec n_2$ meet in a line. The direction of the intersection line is $\vec n_1 \times \vec n_2$ (perpendicular to both normals).

To find a point on the line, set one coordinate to a convenient value (say $z = 0$) and solve the two-by-two linear system in $x$ and $y$.

If $\vec n_1 \parallel \vec n_2$ (parallel planes) the two planes either coincide ($e_1/|\vec n_1| = e_2/|\vec n_2|$) or never meet.

Three planes

Three planes can meet in a point (unique intersection , generic case), in a line (one redundant equation), or in no point (inconsistent). Solve the $3\times 3$ linear system in $(x, y, z)$; the determinant of the coefficient matrix tells you:

• Non-zero $\Rightarrow$ unique point.
• Zero $\Rightarrow$ planes are either parallel or share a common line.

Distance between parallel planes

$Ax + By + Cz = D_1$ and $Ax + By + Cz = D_2$: $D = \frac{|D_1 - D_2|}{\sqrt{A^2 + B^2 + C^2}}.$

Just take any point on one plane and compute its distance to the other.

Worked examples

Example 1. Distance from $(1, 1, 1)$ to $2x + y - z = 5$.

$D = \dfrac{|2 + 1 - 1 - 5|}{\sqrt{4 + 1 + 1}} = \dfrac{3}{\sqrt 6}$.

Example 2. Where does the line $\vec r = (1, 2, 3) + t(2, 1, -1)$ meet the plane $x + y + z = 9$?

$(1 + 2t) + (2 + t) + (3 - t) = 9 \Rightarrow 6 + 2t = 9 \Rightarrow t = 3/2$. Point: $(1 + 3, 2 + 3/2, 3 - 3/2) = (4, 7/2, 3/2)$.

Example 3. Find the line of intersection of $x + y + z = 6$ and $x - y + z = 2$.

Normals: $(1, 1, 1)$ and $(1, -1, 1)$. Cross product: $(2, 0, -2) \propto (1, 0, -1)$. Direction.

A point on the line: set $y = 0$. Then $x + z = 6$ and $x + z = 2$ , inconsistent. So try $z = 0$: $x + y = 6$, $x - y = 2$ gives $x = 4, y = 2$. Point $(4, 2, 0)$.

Line: $\vec r = (4, 2, 0) + t(1, 0, -1)$.

Example 4. Find the distance between the parallel planes $2x - 2y + z = 5$ and $2x - 2y + z = 11$.

$D = \dfrac{|5 - 11|}{\sqrt{4 + 4 + 1}} = \dfrac{6}{3} = 2$.

Example 5. Foot of perpendicular from $(1, 0, 0)$ on the plane $x + y + z = 1$.

Parametrise along normal: $(1 + t, t, t)$. On plane: $1 + 3t = 1$, so $t = 0$. Foot: $(1, 0, 0)$ , the point is on the plane.

Use $(0, 0, 0)$ instead: $(t, t, t)$, $3t = 1$, $t = 1/3$. Foot: $(1/3, 1/3, 1/3)$.

Example 6. Determine whether the planes $x + y + z = 1$, $x - y + z = 3$, $2x + z = 4$ meet in a single point, in a line, or are inconsistent.

Coefficient matrix: $\begin{pmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 2 & 0 & 1 \end{pmatrix}$. Determinant: $1(-1 - 0) - 1(1 - 2) + 1(0 + 2) = -1 + 1 + 2 = 2 \ne 0$. Unique solution.

Solve: from first two, subtract: $2y = -2 \Rightarrow y = -1$. From third: $2x + z = 4$. Substitute into first: $x + z = 2$. From these two: $x = 2$, $z = 0$. So intersection: $(2, -1, 0)$.

Try it yourself

1. Distance from $(0, 0, 0)$ to $x + 2y + 2z = 9$.
2. Distance from $(1, 2, 3)$ to $x - 2y - 2z = 1$.
3. Find where the line $\vec r = (2, 1, 0) + t(1, -1, 1)$ meets the plane $x + y + z = 5$.
4. Line of intersection of $2x + 3y - z = 5$ and $x - y + 2z = 7$.
5. Distance between parallel planes $x + 2y + 3z = 14$ and $x + 2y + 3z = 21$.
6. Foot of perpendicular from $(3, -1, 4)$ to $2x - 3y + z = 7$.
7. Find $\lambda$ if the distance from $(2, 1, 5)$ to $2x + 2y - z = \lambda$ is $3$.
8. Show that the planes $x + y + z = 1$, $2x + 2y + 2z = 3$, $x + y + z = 5$ have no common point.
9. Three planes: $x + y = 2$, $y + z = 4$, $x + z = 6$. Find their common point.
10. Find the line of intersection of the $xy$-plane and the plane $x + y + z = 6$.
11. Find the plane through the line of intersection of $x + 2y - z = 1$ and $2x - y + z = 3$, and through the point $(1, 1, 1)$.
12. Distance from $(1, 1, 1)$ to the line $\vec r = (0, 0, 0) + t(1, 1, 0)$.
13. Show that lines $\vec r = (1, 1, 0) + t(1, 0, 1)$ and $\vec r = (0, 1, 1) + s(1, 1, 0)$ lie in a common plane and find that plane.
14. Find a point equidistant from the planes $2x + y - z = 5$, $x - y + z = 3$, and $x - 2y + z = 1$.

Pitfalls and tricks

• The formula $|Ax_0 + By_0 + Cz_0 + D|/\sqrt{\cdots}$ is the workhorse for point-to-plane distance.
• Line and plane: solve the parametric equation in the plane equation. Set $\vec d\cdot\vec n = 0$ as a special-case check first.
• Two planes meet in a line whose direction is $\vec n_1\times\vec n_2$. Find a point by setting one coordinate.
• For three planes, use determinants. Non-zero $\Rightarrow$ unique point; zero $\Rightarrow$ check consistency separately.
• Always sketch the configuration , even rough , to verify your algebraic answer.