Equations of Planes

A plane is a flat surface that extends forever, like an endless sheet of glass. Just like a line, a plane can be described in several ways: with a normal vector (scalar equation) or with two direction vectors (vector and parametric equations). Being able to switch between them is the key skill for every intersection and distance problem in the rest of this unit.

Key ideas

What fixes a plane?

Any one of these determines exactly one plane:

• a point and a normal vector,
• a point and two non-parallel direction vectors,
• three points that are not on one line (non-collinear),
• a line and a point not on it.

The normal vector

A normal to a plane is a non-zero vector perpendicular to the plane, which means perpendicular to every vector that lies in the plane. A plane’s normal direction never changes from point to point (it’s flat), and any non-zero scalar multiple of a normal is also a normal. So $[3, 2, 4]$, $[6, 4, 8]$ and $[-3, -2, -4]$ are all normals to the same plane.

The scalar equation

Let $P_0(x_0, y_0, z_0)$ be a point on the plane and $\vec{n} = [A, B, C]$ a normal. For any point $P(x, y, z)$ in the plane, $\overrightarrow{P_0P}$ lies in the plane, so $\vec{n} \cdot \overrightarrow{P_0P} = 0$:

$$A(x - x_0) + B(y - y_0) + C(z - z_0) = 0$$

Expanding and collecting the constants into $D$ gives the scalar equation of the plane:

$$Ax + By + Cz + D = 0, \qquad \vec{n} = [A, B, C]$$

You can read a normal straight off the coefficients. For example, a normal to $3x + 2y + 4z - 12 = 0$ is $[3, 2, 4]$. This is the 3-space version of the scalar equation of a line in 2-space.

The plane $3x + 2y + 4z = 12$, sketched from its intercepts, with normal $\vec{n} = [3, 2, 4]$.

Vector and parametric equations

If $\vec{a}$ and $\vec{b}$ are two non-parallel vectors in the plane and $\vec{r}_0$ is the position vector of a point on it, every point of the plane can be reached by going to $P_0$ and then some amount of $\vec{a}$ plus some amount of $\vec{b}$:

$$\vec{r} = \vec{r}_0 + s\vec{a} + t\vec{b}, \qquad s, t \in \mathbb{R}$$

A line needs one parameter; a plane needs two. The parametric equations are the components:

$$x = x_0 + s a_1 + t b_1, \qquad y = y_0 + s a_2 + t b_2, \qquad z = z_0 + s a_3 + t b_3$$

This form also shows a nice property: the sum of any two vectors in a plane ($s\vec{a} + t\vec{b}$) also lies in the plane.

Converting between forms

• Vector to scalar: a normal is perpendicular to both direction vectors, so use the cross product: $\vec{n} = \vec{a} \times \vec{b}$. Then substitute the point to find $D$.
• Scalar to vector: find three points on the plane (the intercepts are often easiest), then use two vectors between them as $\vec{a}$ and $\vec{b}$. Check that each is perpendicular to $\vec{n}$ (dot product $0$).
• Three points to any form: with points $A$, $B$, $C$, use $\vec{a} = \overrightarrow{AB}$ and $\vec{b} = \overrightarrow{AC}$, and $\vec{n} = \overrightarrow{AB} \times \overrightarrow{AC}$.

Parallel and perpendicular planes

Compare the normals $\vec{n}_1$ and $\vec{n}_2$:

Relationship	Test
Parallel	$\vec{n}_1 = k\vec{n}_2$ for some scalar $k$
Coincident (same plane)	the whole equations are multiples of each other
Perpendicular	$\vec{n}_1 \cdot \vec{n}_2 = 0$

Worked examples

Example 1: From a point and a normal

Find the scalar equation of the plane through $P(2, -1, 3)$ with normal $\vec{n} = [4, 1, -2]$.

Solution. The equation has the form $4x + y - 2z + D = 0$. Substitute $P$:

$$4(2) + (-1) - 2(3) + D = 0 \quad\Rightarrow\quad 1 + D = 0 \quad\Rightarrow\quad D = -1$$ $$4x + y - 2z - 1 = 0$$

Example 2: Through three points

Find vector, parametric, and scalar equations of the plane through $A(1, 0, 2)$, $B(3, 1, 0)$ and $C(0, 2, 1)$.

Solution. Two vectors in the plane:

$$\overrightarrow{AB} = [2, 1, -2], \qquad \overrightarrow{AC} = [-1, 2, -1]$$

They aren’t parallel, so the points are non-collinear. Vector equation:

$$\vec{r} = [1, 0, 2] + s[2, 1, -2] + t[-1, 2, -1], \quad s, t \in \mathbb{R}$$

Parametric equations:

$$x = 1 + 2s - t, \qquad y = s + 2t, \qquad z = 2 - 2s - t$$

For the scalar equation, find a normal with the cross product:

$$\begin{aligned} \vec{n} &= [2, 1, -2] \times [-1, 2, -1] \\ &= [1(-1) - (-2)(2),\ (-2)(-1) - 2(-1),\ 2(2) - 1(-1)] \\ &= [3, 4, 5] \end{aligned}$$

So $3x + 4y + 5z + D = 0$. Using $A$: $3 + 0 + 10 + D = 0$, so $D = -13$:

$$3x + 4y + 5z - 13 = 0$$

Check $B$: $9 + 4 + 0 - 13 = 0$ ✓. Check $C$: $0 + 8 + 5 - 13 = 0$ ✓.

Example 3: Vector to scalar

Find the scalar equation of the plane $\vec{r} = [3, -2, 1] + s[1, 2, 0] + t[0, 1, -1]$.

Solution.

$$\vec{n} = [1, 2, 0] \times [0, 1, -1] = [2(-1) - 0(1),\ 0(0) - 1(-1),\ 1(1) - 2(0)] = [-2, 1, 1]$$

So $-2x + y + z + D = 0$. Using $(3, -2, 1)$: $-6 - 2 + 1 + D = 0$, so $D = 7$:

$$-2x + y + z + 7 = 0 \qquad \text{or, multiplying by } -1, \qquad 2x - y - z - 7 = 0$$

Check with $s = t = 1$, the point $(4, 1, 0)$: $8 - 1 - 0 - 7 = 0$ ✓.

Example 4: Scalar to vector

Find a vector equation of the plane $x - 2y + 3z - 6 = 0$.

Solution. Find the intercepts:

• $x$-intercept: $x = 6$, point $(6, 0, 0)$
• $y$-intercept: $-2y = 6$, point $(0, -3, 0)$
• $z$-intercept: $3z = 6$, point $(0, 0, 2)$

Vectors from $(6, 0, 0)$ to the other two points are $[-6, -3, 0]$ and $[-6, 0, 2]$. Simplify them to $[2, 1, 0]$ and $[3, 0, -1]$ (multiplying by $-\dfrac{1}{3}$ and $-\dfrac{1}{2}$).

$$\vec{r} = [6, 0, 0] + s[2, 1, 0] + t[3, 0, -1], \quad s, t \in \mathbb{R}$$

Check that both directions are perpendicular to $\vec{n} = [1, -2, 3]$: $2 - 2 + 0 = 0$ ✓ and $3 + 0 - 3 = 0$ ✓.

Common mistakes

Mixing up a line’s equation and a plane’s. $\vec{r} = \vec{r}_0 + t\vec{m}$ (one parameter) is a line; $\vec{r} = \vec{r}_0 + s\vec{a} + t\vec{b}$ (two parameters) is a plane. And in 3-space, $Ax + By + Cz + D = 0$ is always a plane.

Using parallel direction vectors. If $\vec{a}$ and $\vec{b}$ are parallel, $s\vec{a} + t\vec{b}$ only covers a line. When you use three points, check they are not collinear (their cross product mustn’t be $\vec{0}$).

Cross product slips. Most wrong scalar equations come from a sign error in $\vec{a} \times \vec{b}$. Always check your normal with a dot product against both direction vectors, then check a second point in the final equation.

Forgetting to find D. The normal alone gives a whole family of parallel planes. You need to substitute a point to pick the right one.

Deciding planes are perpendicular because the equations “look different”. Use the test: perpendicular planes have $\vec{n}_1 \cdot \vec{n}_2 = 0$. Planes that are neither parallel nor perpendicular are common.

Practice

1. (Warm-up) State a normal vector to the plane $5x - y + 2z = 8$, and two other normals. Is $(1, 1, 2)$ on the plane?

Solution

A normal is $[5, -1, 2]$. Any non-zero multiple also works, such as $[10, -2, 4]$ and $[-5, 1, -2]$.

$5(1) - 1 + 2(2) = 8$ ✓, so $(1, 1, 2)$ is on the plane.

2. (Warm-up) Find the scalar equation of the plane through $(0, 3, -1)$ with normal $[2, -5, 1]$.

Solution

$2x - 5y + z + D = 0$. Substitute: $0 - 15 - 1 + D = 0$, so $D = 16$.

$$2x - 5y + z + 16 = 0$$

3. (Warm-up) Find the scalar equation of the plane through the origin parallel to $3x + y - 4z = 7$.

Solution

Parallel planes can share a normal, $[3, 1, -4]$. Through the origin, $D = 0$:

$$3x + y - 4z = 0$$

4. (Core) For the plane $\vec{r} = [0, 1, 4] + s[2, 1, -1] + t[1, 3, 0]$, write the parametric equations and the scalar equation.

Solution

Parametric: $x = 2s + t$, $y = 1 + s + 3t$, $z = 4 - s$.

Normal:

$$[2, 1, -1] \times [1, 3, 0] = [1(0) - (-1)(3),\ (-1)(1) - 2(0),\ 2(3) - 1(1)] = [3, -1, 5]$$

So $3x - y + 5z + D = 0$. Using $(0, 1, 4)$: $0 - 1 + 20 + D = 0$, so $D = -19$:

$$3x - y + 5z - 19 = 0$$

Check with $s = 1$, $t = 0$, the point $(2, 2, 3)$: $6 - 2 + 15 - 19 = 0$ ✓.

5. (Core) Find the scalar equation of the plane through $P(2, 1, -1)$, $Q(4, 3, 0)$ and $R(1, 0, 2)$. What do you notice about this plane?

Solution

$\overrightarrow{PQ} = [2, 2, 1]$ and $\overrightarrow{PR} = [-1, -1, 3]$.

$$\overrightarrow{PQ} \times \overrightarrow{PR} = [2(3) - 1(-1),\ 1(-1) - 2(3),\ 2(-1) - 2(-1)] = [7, -7, 0]$$

Use the simpler normal $[1, -1, 0]$: $x - y + D = 0$. With $P$: $2 - 1 + D = 0$, so $D = -1$:

$$x - y - 1 = 0$$

Check $Q$: $4 - 3 - 1 = 0$ ✓. Check $R$: $1 - 0 - 1 = 0$ ✓.

There’s no $z$ term, so the plane is parallel to the $z$-axis (a vertical “wall”).

6. (Core) Decide whether each pair of planes is parallel, perpendicular, or neither.

• (a) $2x - y + 3z = 5$ and $-4x + 2y - 6z = 1$
• (b) $x + 2y - z = 3$ and $3x - y + z = 0$
• (c) $x + y + z = 1$ and $x - y + 2z = 4$

Solution

(a) $[-4, 2, -6] = -2[2, -1, 3]$, so the normals are parallel. Multiplying the first equation by $-2$ gives a right side of $-10$, not $1$, so the planes are parallel and distinct.

(b) $[1, 2, -1] \cdot [3, -1, 1] = 3 - 2 - 1 = 0$, so the planes are perpendicular.

(c) The normals $[1, 1, 1]$ and $[1, -1, 2]$ aren’t multiples, and their dot product is $1 - 1 + 2 = 2 \ne 0$. Neither.

7. (Core) Find a vector equation of the plane $2x + y - z = 4$.

Solution

Intercepts: $(2, 0, 0)$, $(0, 4, 0)$ and $(0, 0, -4)$.

From $(2, 0, 0)$, the vectors to the other two are $[-2, 4, 0]$ and $[-2, 0, -4]$. Simplify to $[-1, 2, 0]$ and $[1, 0, 2]$:

$$\vec{r} = [2, 0, 0] + s[-1, 2, 0] + t[1, 0, 2]$$

Check against $\vec{n} = [2, 1, -1]$: $-2 + 2 + 0 = 0$ ✓ and $2 + 0 - 2 = 0$ ✓.

8. (Challenge) Find the scalar equation of the plane that contains the line $\vec{r} = [1, 2, 0] + t[1, -1, 2]$ and the point $P(3, 0, 1)$.

Solution

One direction vector is the line’s, $\vec{a} = [1, -1, 2]$. A second goes from the point $(1, 2, 0)$ on the line to $P$: $\vec{b} = [2, -2, 1]$.

$$\vec{a} \times \vec{b} = [(-1)(1) - 2(-2),\ 2(2) - 1(1),\ 1(-2) - (-1)(2)] = [3, 3, 0]$$

Use $\vec{n} = [1, 1, 0]$: $x + y + D = 0$. With $(1, 2, 0)$: $D = -3$.

$$x + y - 3 = 0$$

Check $P$: $3 + 0 - 3 = 0$ ✓. Check the line point at $t = 1$, $(2, 1, 2)$: $2 + 1 - 3 = 0$ ✓.

9. (Challenge) Find the scalar equation of the plane through $A(1, -2, 3)$ that is perpendicular to both planes $x + y - z = 2$ and $2x - y + z = 5$.

Solution

If the new plane is perpendicular to a plane, its normal is perpendicular to that plane’s normal. So the new normal is perpendicular to both $[1, 1, -1]$ and $[2, -1, 1]$:

$$[1, 1, -1] \times [2, -1, 1] = [1(1) - (-1)(-1),\ (-1)(2) - 1(1),\ 1(-1) - 1(2)] = [0, -3, -3]$$

Use $\vec{n} = [0, 1, 1]$: $y + z + D = 0$. With $A$: $-2 + 3 + D = 0$, so $D = -1$:

$$y + z - 1 = 0$$

Check: $[0, 1, 1] \cdot [1, 1, -1] = 0$ ✓ and $[0, 1, 1] \cdot [2, -1, 1] = 0$ ✓.