Planes through lines and points

What you will be able to do

• write every plane through the line where two planes meet, and pick the one that passes through a point or is perpendicular to another plane;
• find the plane through a line and a point, and the plane through two lines;
• choose between these methods.

Pages of a book

Hold a book half open. The spine is a line. Every page is a plane, and all the pages share the spine. Turning a page about the spine gives the next plane of the family (or pencil) of planes through that line. Two numbers name one page: its angle. In algebra the one number is $\lambda$.

The family through the line of two planes

Let two planes be $P_1 \equiv A_1x + B_1y + C_1z - k_1 = 0$ and $P_2 \equiv A_2x + B_2y + C_2z - k_2 = 0$. They meet in a line. Every point of that line makes both expressions $0$. So it also makes $P_1 + \lambda P_2$ equal to $0$, for every value of $\lambda$. The equation

is a plane for each $\lambda$, and it contains the common line. Every plane through the line is of this kind, except $P_2$ itself, which belongs to $\lambda \to \infty$. We may write $\lambda P_1 + \mu P_2 = 0$ to include both.

To pick one plane we need one more fact:

• A point $(x_0, y_0, z_0)$ on the plane. Put it in: $P_1(x_0, y_0, z_0) + \lambda P_2(x_0, y_0, z_0) = 0$.
• Perpendicular to a plane with normal $\vec{n}_3$. The normal $\vec{n}_1 + \lambda\vec{n}_2$ must satisfy $(\vec{n}_1 + \lambda\vec{n}_2) \cdot \vec{n}_3 = 0$.
• Parallel to a line with direction $\vec{b}$. The normal must satisfy $(\vec{n}_1 + \lambda\vec{n}_2) \cdot \vec{b} = 0$.
• Equal intercepts on two axes, or a given intercept. Collect the terms and compare the coefficients.

Through a line and a point, or through two lines

A line and a point. Take the line $\vec{r} = \vec{a} + t\vec{b}$ and a point $P$ that is not on it. The plane contains $\vec{b}$ and also $\overrightarrow{AP}$. A normal is $\overrightarrow{AP} \times \vec{b}$.

Two lines. If they intersect, the plane contains both directions, so a normal is $\vec{b}_1 \times \vec{b}_2$. If they are parallel, use $(\vec{a}_2 - \vec{a}_1) \times \vec{b}$. Skew lines do not lie in a plane. The plane through one line parallel to another also has the normal $\vec{b}_1 \times \vec{b}_2$.

In every case, find the constant by putting a known point in the equation, and then check a second point.

A worked example

The family. The planes $x + y + z = 6$ and $2x - y + z = 3$ meet in a line. Find the plane through that line and the point $(1, 1, 0)$.

1. The family. $(x + y + z - 6) + \lambda(2x - y + z - 3) = 0$.
2. The point. At $(1, 1, 0)$ the expressions are $-4$ and $-2$. So $-4 - 2\lambda = 0$ and $\lambda = -2$.
3. The plane. $(x + y + z - 6) - 2(2x - y + z - 3) = -3x + 3y - z = 0$, which is $3x - 3y + z = 0$.
4. A check. The point $(3, 3, 0)$ is on both given planes: $3 + 3 + 0 = 6$ and $6 - 3 + 0 = 3$. It satisfies $9 - 9 + 0 = 0$.

A line and a point. Find the plane through the line $\dfrac{x - 2}{1} = \dfrac{y + 1}{2} = \dfrac{z}{2}$ and the point $(1, 1, 1)$.

1. A normal. $\overrightarrow{AP} = (-1, 2, 1)$ with $A = (2, -1, 0)$, and $\vec{b} = (1, 2, 2)$. Then $\overrightarrow{AP} \times \vec{b} = (2, 3, -4)$.
2. The plane. $2x + 3y - 4z = 2 + 3 - 4 = 1$.
3. The check. $A$ gives $4 - 3 - 0 = 1$.

Practise with new numbers

Try it in Python

The program finds the member of a family through a point and checks it on a point of the common line.

Common mistakes

• Writing the family with the constants on one side only. Both expressions must be written as $A x + B y + C z - k$ before you add them.
• Losing a plane. $P_2$ itself is not in $P_1 + \lambda P_2 = 0$ for any finite $\lambda$. Check it separately.
• Using a point on the common line to find $\lambda$. Every $\lambda$ works for such a point, so it tells you nothing.
• Taking the cross product in the wrong order. The sign of the normal changes, but the plane stays the same, as long as you find the constant from a point.

Remember these

• Planes through the line of $P_1 = 0$ and $P_2 = 0$: $P_1 + \lambda P_2 = 0$.
• Fix $\lambda$ with a point, a perpendicularity condition or an intercept condition.
• Plane through a line and a point: normal $\overrightarrow{AP} \times \vec{b}$.
• Plane through two lines: normal $\vec{b}_1 \times \vec{b}_2$, or $(\vec{a}_2 - \vec{a}_1) \times \vec{b}$ for parallel lines.

Check yourself

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

1. Write the family of planes through the line of intersection of $x - y = 0$ and $z - 2 = 0$.

2. Find the member of the family $(x + y + z - 6) + \lambda(2x - y + z - 3) = 0$ that passes through the origin.

3. The planes $x + y + z = 3$ and $2x - y + 3z = 2$ are given. Find the member of the family that is perpendicular to the plane $2x + y = 7$.

4. Find a normal to the plane through the line $\dfrac{x}{1} = \dfrac{y}{1} = \dfrac{z}{1}$ and the point $(1, 0, 0)$.

5. Why can no plane contain two skew lines?

References

• OpenStax, Calculus Volume 3, Chapter 2, section on equations of lines and planes in space.
• NCERT, Mathematics Part II for Class XII, Chapter 11, exercises on planes through the intersection of two planes.