The Plane

A plane is a flat, two-dimensional surface that extends without bound. In three dimensions a plane is pinned down the moment you fix its orientation (which way it faces) and one point it passes through. Orientation is captured by a single vector $\vec n$ that is perpendicular to the plane, called the normal. Almost every plane formula in this chapter is just a restatement of one fact: a point lies on the plane exactly when the vector from a fixed point of the plane to it is perpendicular to $\vec n$.

Equation of a plane in normal form

Let $\hat n$ be the unit normal pointing from the origin towards the plane, and let $d$ be the perpendicular distance of the plane from the origin. A point with position vector $\vec r$ lies on the plane iff its projection on $\hat n$ equals $d$:

If $\hat n=(l,m,n)$ are its direction cosines (so $l^2+m^2+n^2=1$), the Cartesian normal form is

Here $l,m,n$ are direction cosines of the normal, not of any line in the plane, and $d$ is genuinely the distance from the origin only when the left side is normalised.

Plane through a point with a given normal

If the plane passes through the point $\vec a$ and has normal $\vec N$, then for any point $\vec r$ on it the vector $\vec r-\vec a$ lies in the plane and so is perpendicular to $\vec N$:

In Cartesian form, with $\vec a=(x_1,y_1,z_1)$ and $\vec N=(A,B,C)$,

General equation and its normal

Expanding the previous line gives the general first-degree equation

Every first-degree equation in $x,y,z$ (with $A,B,C$ not all zero) is a plane, and the coefficients read off the normal directly: $\vec N=(A,B,C)$. This single observation drives angle and perpendicularity questions, because comparing planes reduces to comparing their normals.

Intercept form

If a plane meets the axes at $(a,0,0)$, $(0,b,0)$, $(0,0,c)$ then it can be written

where $a,b,c$ are the $x$-, $y$- and $z$-intercepts. To find intercepts from $Ax+By+Cz+D=0$, set the other two variables to zero, e.g. $a=-D/A$.

Plane through three points

Three non-collinear points $\vec a,\vec b,\vec c$ determine a plane. A normal is $(\vec b-\vec a)\times(\vec c-\vec a)$, and the condition that $\vec r$ is coplanar with them is the scalar triple product

In coordinates this is the determinant

Plane through the intersection of two planes

If $P_1\equiv \vec r\cdot\vec n_1-d_1=0$ and $P_2\equiv \vec r\cdot\vec n_2-d_2=0$, then for every scalar $\lambda$

is a plane containing their entire line of intersection. Choose $\lambda$ to satisfy one extra condition (a point it must pass through, or a normal it must be perpendicular/parallel to).

Angle between two planes

The angle between two planes equals the angle between their normals $\vec n_1,\vec n_2$:

The planes are perpendicular iff $\vec n_1\cdot\vec n_2=0$ and parallel iff $\vec n_1\times\vec n_2=\vec 0$ (proportional coefficients). The absolute value again picks the acute angle.

Angle between a line and a plane

A line with direction $\vec b$ makes an angle $\theta$ with a plane of normal $\vec n$. Since $\vec b$ makes the complementary angle with $\vec n$,

The line is parallel to the plane iff $\vec b\cdot\vec n=0$ (and the line is not in the plane) and perpendicular to the plane iff $\vec b\parallel\vec n$.

Distance of a point from a plane

The perpendicular distance of $(x_1,y_1,z_1)$ from $Ax+By+Cz+D=0$ is

In vector form, the distance of $\vec a$ from $\vec r\cdot\hat n=d$ is $|\vec a\cdot\hat n-d|$. Putting $(x_1,y_1,z_1)=(0,0,0)$ recovers the distance of the plane from the origin.

Foot of the perpendicular and image of a point

To drop a perpendicular from $P(x_1,y_1,z_1)$ to $Ax+By+Cz+D=0$, move from $P$ along the normal direction $(A,B,C)$. Writing the foot as $P+t(A,B,C)$ and substituting,

The foot $Q$ is $P+t(A,B,C)$; the image (mirror reflection) $P'$ is $P+2t(A,B,C)$, since the foot is the midpoint of $P$ and its image.

Intersection of a line and a plane

Substitute the line's parametric point $\vec a+\lambda\vec b$ into the plane equation and solve for the single parameter $\lambda$; back-substitution gives the point of intersection. If $\vec b\cdot\vec n=0$ the line is parallel to the plane — there is either no solution (line outside) or every $\lambda$ works (line lies in the plane).

Coplanarity of two lines

Lines $\vec r=\vec a_1+\lambda\vec b_1$ and $\vec r=\vec a_2+\mu\vec b_2$ are coplanar iff the vector joining a point of each is perpendicular to $\vec b_1\times\vec b_2$:

equivalently the determinant of $(\vec a_2-\vec a_1)$, $\vec b_1$, $\vec b_2$ vanishes. Coplanar non-parallel lines intersect; the plane that holds them has normal $\vec b_1\times\vec b_2$.