Determine the vector equation and the Cartesian equation of a plane from a point and a normal or from three points, find the distance from a point to a plane, and find the line of intersection or angle between planes

What this dot point is asking

QCAA wants you to describe a plane in three dimensions using a normal vector, write it in both vector and Cartesian form, build it from three points, and answer geometric questions: the distance from a point to the plane and the angle between two planes. The normal vector is the central object, since a plane is fixed by one point on it and a direction perpendicular to it.

The answer

Normal vector and the vector equation

A plane is determined by a point $A$ with position vector $\mathbf{a}$ and a normal vector $\mathbf{n}$ perpendicular to the plane. A point $P$ with position vector $\mathbf{r}$ lies on the plane exactly when $\vec{AP}$ is perpendicular to $\mathbf{n}$, that is $(\mathbf{r} - \mathbf{a})\cdot\mathbf{n} = 0$. This rearranges to

Cartesian equation

Writing $\mathbf{n} = (a, b, c)$ and $\mathbf{r} = (x, y, z)$, the vector equation becomes

equivalently $ax + by + cz + d' = 0$. The coefficients of $x$, $y$ and $z$ are precisely the components of the normal vector, which is how you read a normal straight off a Cartesian equation.

Plane through three points

Given three non-collinear points $A$, $B$, $C$, form two vectors in the plane, $\vec{AB}$ and $\vec{AC}$. Their cross product is normal to both, hence normal to the plane:

Then use $\mathbf{r}\cdot\mathbf{n} = \mathbf{a}\cdot\mathbf{n}$ with any one of the three points.

Distance from a point to a plane

The perpendicular distance from a point $P_0 = (x_0, y_0, z_0)$ to the plane $ax + by + cz + d' = 0$ is

The absolute value gives an unsigned distance, and the denominator is the magnitude of the normal.

Angle between two planes

The angle between two planes equals the angle between their normals. For normals $\mathbf{n}_1$ and $\mathbf{n}_2$,

The absolute value selects the acute angle between the planes. Two planes are parallel when their normals are scalar multiples.

Line of intersection

Two non-parallel planes meet in a line whose direction is $\mathbf{n}_1 \times \mathbf{n}_2$, since the line lies in both planes and so is perpendicular to both normals. Finding one common point fixes the line. To locate a point, set one variable to a convenient value (often $z = 0$) and solve the two Cartesian equations simultaneously for the other two coordinates. The point plus the direction $\mathbf{n}_1 \times \mathbf{n}_2$ then gives the vector equation of the line of intersection.

Reading the geometry from the equation

The Cartesian form $ax + by + cz = d$ packs the whole geometry into four numbers. The vector $(a, b, c)$ points perpendicular to the plane, so two planes are parallel exactly when their coefficient triples are proportional, and identical when the constants are proportional in the same ratio too. If $d = 0$ the plane passes through the origin. Dividing through by $|\mathbf{n}| = \sqrt{a^2 + b^2 + c^2}$ produces the normal form, in which the right-hand side is the signed distance from the origin to the plane, a useful check on distance problems.

Angle between a line and a plane

A related calculation QCAA sometimes asks for is the angle between a line of direction $\mathbf{d}$ and a plane of normal $\mathbf{n}$. Because the normal is perpendicular to the plane, the angle $\alpha$ between the line and the plane is the complement of the angle between $\mathbf{d}$ and $\mathbf{n}$:

Note the sine, not cosine, because we measure from the plane rather than from the normal. A line is parallel to a plane when $\mathbf{d} \cdot \mathbf{n} = 0$, and perpendicular to it when $\mathbf{d}$ is a scalar multiple of $\mathbf{n}$.