Equations of a plane

What you will be able to do

• write the equation of a plane from a point and a normal, in vector and Cartesian form;
• find the plane through three points and the plane with given intercepts;
• read the normal and the distance from the origin from an equation.

A roof, a wall and a sheet of glass

A flat roof tilts at some angle. A pole standing straight up from the roof is at right angles to every line in the roof. The pole gives the normal of the roof, and the roof is a plane. One point of the roof and the direction of the pole fix the roof completely.

The vector equation

Let $\vec{a}$ be the position vector of a known point $A$ of the plane, and $\vec{n}$ a normal. A point $P$ with position vector $\vec{r}$ lies in the plane exactly when $\overrightarrow{AP} = \vec{r} - \vec{a}$ is perpendicular to $\vec{n}$:

If $\hat{n}$ is the unit normal, the equation is $\vec{r} \cdot \hat{n} = d$, and $d \ge 0$ is the distance of the plane from the origin. This is the normal form of the Indian textbooks. A-Level and IB books write $\mathbf{r} \cdot \mathbf{n} = p$ with any normal, and $\dfrac{p}{\lvert \mathbf{n} \rvert}$ is the distance from the origin.

The Cartesian equation

Put $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ and $\vec{n} = A\hat{i} + B\hat{j} + C\hat{k}$. The equation becomes

The coefficients $(A, B, C)$ are the components of a normal. The distance of the plane from the origin is $\dfrac{\lvert D \rvert}{\sqrt{A^2 + B^2 + C^2}}$.

Other ways to fix a plane

• Through three points $P$, $Q$, $R$ that are not on a line. A normal is $\overrightarrow{PQ} \times \overrightarrow{PR}$. Use any one of the points to find the constant.
• With intercepts $a$, $b$, $c$ on the axes: $\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 1$.
• Through a point $\vec{a}$ parallel to two vectors $\vec{b}$ and $\vec{c}$: $\vec{r} = \vec{a} + \lambda\vec{b} + \mu\vec{c}$. A normal is $\vec{b} \times \vec{c}$.
• Special planes. $x = k$ is parallel to the $yz$-plane. The plane $z = 0$ is the $xy$-plane.

A worked example

A point and a normal. Find the plane through $A(1, 2, 3)$ with the normal $2\hat{i} - \hat{j} + 2\hat{k}$.

1. The constant. $\vec{a} \cdot \vec{n} = 2 - 2 + 6 = 6$.
2. The plane. $2x - y + 2z = 6$.
3. The normal form. $\lvert \vec{n} \rvert = 3$, so $\vec{r} \cdot \left(\dfrac{2}{3}\hat{i} - \dfrac{1}{3}\hat{j} + \dfrac{2}{3}\hat{k}\right) = 2$. The plane is $2$ units from the origin.

Three points. Find the plane through $(1, 1, 1)$, $(2, 3, 4)$ and $(0, 1, 3)$.

1. Two vectors in the plane. $(1, 2, 3)$ and $(-1, 0, 2)$.
2. A normal. $(1, 2, 3) \times (-1, 0, 2) = (4, -5, 2)$.
3. The plane. $4x - 5y + 2z = 4 - 5 + 2 = 1$. Check with $(2, 3, 4)$: $8 - 15 + 8 = 1$.

Intercepts. The plane through $(1, 0, 0)$, $(0, 2, 0)$ and $(0, 0, 3)$ is $x + \dfrac{y}{2} + \dfrac{z}{3} = 1$, which is $6x + 3y + 2z = 6$.

Practise with new numbers

Try it in Python

The program finds the plane through three points from a cross product, then checks that all three points satisfy the equation.

Common mistakes

• Taking the normal from the wrong vectors. The normal comes from the cross product of two vectors that lie in the plane, not from the position vectors of the points.
• Forgetting the constant. The normal gives the coefficients. The point gives the constant $k = \vec{a} \cdot \vec{n}$.
• Reading $-D$ as $D$. In $Ax + By + Cz + D = 0$ the equation is $Ax + By + Cz = -D$.
• Using three points on a line. Then $\overrightarrow{PQ} \times \overrightarrow{PR} = \vec{0}$, and there is no single plane.

Remember these

• $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$, and in coordinates $Ax + By + Cz = k$ with normal $(A, B, C)$.
• The normal form $\vec{r} \cdot \hat{n} = d$ has $d$ equal to the distance from the origin.
• Through three points: a normal is $\overrightarrow{PQ} \times \overrightarrow{PR}$.
• Intercept form: $\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 1$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Find the plane through $(2, -1, 0)$ with normal $3\hat{i} + 4\hat{j} - 12\hat{k}$, and its distance from the origin.

2. Write the plane $2x - 3y + 6z = 14$ in the form $\vec{r} \cdot \hat{n} = d$.

3. A plane meets the axes at $(4, 0, 0)$, $(0, 2, 0)$ and $(0, 0, 4)$. Find its equation.

4. Find a normal to the plane through $(0, 0, 0)$, $(1, 2, 0)$ and $(0, 1, 3)$.

5. The plane $x + y + z = 6$ is given. Is the point $(2, 5, -1)$ on it?

References

• OpenStax, Calculus Volume 3, Chapter 2, section on equations of lines and planes in space (scalar and vector equations of a plane).
• NCERT, Mathematics Part II for Class XII, Chapter 11, section on the plane.