Vector equations of lines and planes in three dimensions, their parametric and Cartesian forms, the use of a direction vector for a line and a normal vector for a plane, and the determination of intersections and the angle between a line and a plane

What this dot point is asking

VCAA wants you to write the vector equation of a line from a point and a direction, and of a plane from a point and a normal, to convert between vector, parametric and Cartesian forms, and to find intersections of lines with planes and the angles between them. This builds directly on the dot and cross products.

The vector equation of a line

A line is fixed by one point on it and a direction. If $\mathbf{a}$ is the position vector of a known point and $\mathbf{d}$ is a direction vector along the line, then every point on the line has position vector

Writing $\mathbf{r} = (x, y, z)$, $\mathbf{a} = (a_1, a_2, a_3)$, $\mathbf{d} = (d_1, d_2, d_3)$ gives the parametric form $x = a_1 + t d_1$, $y = a_2 + t d_2$, $z = a_3 + t d_3$. Eliminating $t$ from these (when no component of $\mathbf{d}$ is zero) gives the symmetric Cartesian form $\frac{x - a_1}{d_1} = \frac{y - a_2}{d_2} = \frac{z - a_3}{d_3}$.

The equation of a plane

A plane is fixed by one point on it and a normal vector $\mathbf{n}$ perpendicular to it. A point $\mathbf{r}$ lies on the plane exactly when the displacement $\mathbf{r} - \mathbf{a}$ is perpendicular to $\mathbf{n}$:

In coordinates with $\mathbf{n} = (n_1, n_2, n_3)$ this is the Cartesian equation

The coefficients of $x, y, z$ are the components of a normal vector, which is how you read a normal straight off a Cartesian plane equation.

Intersections and angles

Line meets plane. Substitute the parametric coordinates of the line into the plane's Cartesian equation. This gives one equation in $t$; solve it, then back-substitute to get the intersection point. If the coefficient of $t$ is zero, the line is parallel to the plane (no intersection, or the line lies in the plane).

Angle between a line and a plane. The direction $\mathbf{d}$ and normal $\mathbf{n}$ give the angle $\phi$ between line and plane via

The sine appears (not cosine) because the angle to the plane is the complement of the angle to its normal.

Examples in context

Example 1. The plane through $(0, 0, 0)$ with normal $(1, 1, 1)$ is $x + y + z = 0$.

Example 2. The line through $(1, 0, 0)$ and $(1, 0, 2)$ has direction $(0, 0, 2)$, so $\mathbf{r} = (1, 0, 0) + t(0, 0, 1)$.

Try this

Q1. Write the vector equation of the line through $(2, 1, 5)$ with direction $(1, 0, -2)$. [2 marks]

• Cue. $\mathbf{r} = (2, 1, 5) + t(1, 0, -2)$.

Q2. State a normal vector to the plane $3x - y + 4z = 7$. [1 mark]

• Cue. $(3, -1, 4)$.

Q3. Does the point $(1, 1, 1)$ lie on the plane $2x + y - z = 2$? [2 marks]

• Cue. $2 + 1 - 1 = 2$, yes.