Intersections of Lines and Planes

Two lines in a plane either cross, run parallel, or are the same line. In 3-space there’s a fourth option: lines that aren’t parallel but still never meet. This page sorts out every way two lines, or a line and a plane, can sit in 3-space, and shows you how to find the intersection point when there is one.

Key ideas

Two lines in 3-space

Two lines in 3-space are always in exactly one of these four situations:

Configuration	Directions parallel?	Common points
Intersecting	no	exactly one
Skew	no	none
Parallel and distinct	yes	none
Coincident	yes	all of them (the same line)

Skew lines are not parallel and don’t intersect. Think of a road running east–west and a highway overpass running north–south above it. They can’t happen in 2-space, because two non-parallel lines in a plane always cross. Skew lines always lie in two parallel planes.

Skew lines: different directions, no common point. They lie in parallel planes.

Testing two lines

Write the lines with different parameters, say $\vec{r} = \vec{r}_1 + t\vec{m}_1$ and $\vec{r} = \vec{r}_2 + s\vec{m}_2$.

1. Compare the directions. Is $\vec{m}_1$ a scalar multiple of $\vec{m}_2$?
2. If they’re parallel: test whether a point of one line is on the other. Yes means coincident; no means parallel and distinct.
3. If they’re not parallel: set the parametric equations equal, component by component. That gives three equations in two unknowns, $t$ and $s$. Solve two of them, then check the third.
   • The third equation works: the lines intersect. Substitute $t$ (or $s$) to get the point.
   • The third equation fails: the lines are skew.

A line and a plane

A line and a plane in 3-space can:

• meet in one point (the line crosses the plane),
• be parallel and distinct (no common points), or
• coincide: the line lies in the plane (infinitely many common points).

To find out which, substitute the line’s parametric equations into the plane’s scalar equation and solve for $t$:

Result of solving for t	Meaning
one value of $t$	one point of intersection; substitute $t$ into the line
a false statement like $0 = 5$	the line is parallel to the plane, no intersection
a true statement like $0 = 0$	every $t$ works: the line lies in the plane

Quick test with vectors: the line is parallel to the plane (or in it) exactly when its direction is perpendicular to the plane’s normal, $\vec{m} \cdot \vec{n} = 0$. If $\vec{m} \cdot \vec{n} \ne 0$, the line crosses the plane at exactly one point.

If the plane is given in vector or parametric form, convert it to a scalar equation first (see equations of planes). It makes the substitution much simpler.

Worked examples

Example 1: Two lines that intersect

Show that these lines intersect, and find the point of intersection.

$$L_1: \vec{r} = [1, 2, 3] + t[1, -1, 2] \qquad L_2: \vec{r} = [2, -2, 8] + s[1, 2, -1]$$

Solution. The directions $[1, -1, 2]$ and $[1, 2, -1]$ aren’t multiples, so the lines aren’t parallel. Set the components equal:

$$\begin{aligned} 1 + t &= 2 + s && (x) \\ 2 - t &= -2 + 2s && (y) \\ 3 + 2t &= 8 - s && (z) \end{aligned}$$

Add the $x$ and $y$ equations: $3 = 3s$, so $s = 1$. Then the $x$ equation gives $t = 2$.

Check the $z$ equation: $3 + 2(2) = 7$ and $8 - 1 = 7$ ✓.

All three equations work, so the lines intersect. With $t = 2$ in $L_1$: $(1 + 2,\ 2 - 2,\ 3 + 4) = (3, 0, 7)$.

Check with $s = 1$ in $L_2$: $(2 + 1,\ -2 + 2,\ 8 - 1) = (3, 0, 7)$ ✓.

Example 2: Skew or coincident?

Let $L_1: \vec{r} = [1, 0, 2] + t[2, 1, -1]$.

• (a) Classify $L_1$ and $L_2: \vec{r} = [0, 4, 1] + s[1, -1, 2]$.
• (b) Classify $L_1$ and $L_3: \vec{r} = [5, 2, 0] + u[-4, -2, 2]$.

Solution.

(a) $[2, 1, -1]$ and $[1, -1, 2]$ aren’t multiples, so the lines aren’t parallel. Set them equal:

$$1 + 2t = s \quad (x), \qquad t = 4 - s \quad (y), \qquad 2 - t = 1 + 2s \quad (z)$$

Substitute the $x$ equation into the $y$ equation: $t = 4 - (1 + 2t)$, so $3t = 3$ and $t = 1$. Then $s = 3$.

Check the $z$ equation: $2 - 1 = 1$, but $1 + 2(3) = 7$. They don’t match, so there’s no common point. The lines are skew.

(b) $[-4, -2, 2] = -2[2, 1, -1]$, so the lines are parallel. Is $(5, 2, 0)$ on $L_1$? From $x$: $1 + 2t = 5$ gives $t = 2$. Then $y = 2$ ✓ and $z = 2 - 2 = 0$ ✓. So the lines are coincident: they’re the same line written two ways.

Example 3: A line and a plane, three ways

Find the intersection of each line with the plane $x + y + z = 7$.

• (a) $\vec{r} = [2, -1, 0] + t[1, 3, -2]$
• (b) $\vec{r} = [1, 1, 1] + t[2, -1, -1]$
• (c) $\vec{r} = [3, 2, 2] + t[2, -1, -1]$

Solution.

(a) Substitute $x = 2 + t$, $y = -1 + 3t$, $z = -2t$:

$$(2 + t) + (-1 + 3t) + (-2t) = 7 \quad\Rightarrow\quad 1 + 2t = 7 \quad\Rightarrow\quad t = 3$$

The point is $(2 + 3,\ -1 + 9,\ -6) = (5, 8, -6)$. Check: $5 + 8 - 6 = 7$ ✓.

(b) Substitute $x = 1 + 2t$, $y = 1 - t$, $z = 1 - t$:

$$(1 + 2t) + (1 - t) + (1 - t) = 7 \quad\Rightarrow\quad 3 = 7$$

That’s false for every $t$, so there’s no intersection: the line is parallel to the plane. (Check: $[2, -1, -1] \cdot [1, 1, 1] = 0$.)

(c) Substitute $x = 3 + 2t$, $y = 2 - t$, $z = 2 - t$:

$$(3 + 2t) + (2 - t) + (2 - t) = 7 \quad\Rightarrow\quad 7 = 7$$

True for every $t$, so every point of the line is on the plane: the line lies in the plane.

Example 4: The foot of a perpendicular

Find the point on the plane $2x - y + 2z - 1 = 0$ closest to $A(4, -1, 5)$, and the distance from $A$ to the plane.

Solution. The closest point is where the line through $A$ perpendicular to the plane meets it. That line runs in the direction of the normal $[2, -1, 2]$:

$$\vec{r} = [4, -1, 5] + t[2, -1, 2]$$

Substitute $x = 4 + 2t$, $y = -1 - t$, $z = 5 + 2t$ into the plane:

$$\begin{aligned} 2(4 + 2t) - (-1 - t) + 2(5 + 2t) - 1 &= 0 \\ 8 + 4t + 1 + t + 10 + 4t - 1 &= 0 \\ 9t + 18 &= 0 \\ t &= -2 \end{aligned}$$

The foot of the perpendicular is $(4 - 4,\ -1 + 2,\ 5 - 4) = (0, 1, 1)$. Check: $0 - 1 + 2 - 1 = 0$ ✓.

The distance is the length of the segment from $A$ to $(0, 1, 1)$:

$$\sqrt{(4 - 0)^2 + (-1 - 1)^2 + (5 - 1)^2} = \sqrt{16 + 4 + 16} = 6$$

You’ll see a faster formula for this distance in distances in 3-space.

Common mistakes

Using the same parameter for both lines. Writing both lines with $t$ asks whether they’re at the same point for the same $t$, which is a different (and usually wrong) question. Use $t$ for one line and $s$ for the other.

Not checking the third equation. Two of the three component equations can almost always be solved. The third one decides between “intersecting” and “skew”. Skip it and you’ll call skew lines intersecting.

Calling non-parallel lines intersecting. That’s true in 2-space but not in 3-space. Non-parallel lines in 3-space can be skew.

Misreading 0 = 0 and 0 = 5. When $t$ cancels out, a true statement means the line lies in the plane; a false statement means no intersection. Neither means "$t = 0$".

Stopping at the value of t. The question asks for a point. Substitute $t$ back into the line’s equations, then check the point in the plane.

Practice

1. (Warm-up) $L_1$ passes through $(1, 0, 2)$ with direction $[2, -4, 6]$, and $L_2$ passes through $(3, -4, 8)$ with direction $[-1, 2, -3]$. Are the lines parallel? Are they coincident?

Solution

$[2, -4, 6] = -2[-1, 2, -3]$, so the lines are parallel.

Is $(3, -4, 8)$ on $L_1$? From $x$: $1 + 2t = 3$ gives $t = 1$. Then $y = 0 - 4 = -4$ ✓ and $z = 2 + 6 = 8$ ✓. Yes, so the lines are coincident.

2. (Warm-up) Does the line $\vec{r} = [1, 2, 3] + t[1, 0, -1]$ intersect the plane $x + z = 5$?

Solution

Substitute: $(1 + t) + (3 - t) = 5$ gives $4 = 5$, which is false. No intersection: the line is parallel to the plane. (Check: $[1, 0, -1] \cdot [1, 0, 1] = 0$.)

3. (Core) Find the point where the line $x = 3 + t$, $y = -2 + 2t$, $z = 1 - t$ meets the plane $2x + y - 3z = 15$.

Solution

$$\begin{aligned} 2(3 + t) + (-2 + 2t) - 3(1 - t) &= 15 \\ 6 + 2t - 2 + 2t - 3 + 3t &= 15 \\ 7t + 1 &= 15 \\ t &= 2 \end{aligned}$$

The point is $(5, 2, -1)$. Check: $10 + 2 + 3 = 15$ ✓.

4. (Core) Show that $L_1: \vec{r} = [2, 1, 0] + t[1, 1, 1]$ and $L_2: \vec{r} = [-1, 4, -1] + s[2, -1, 1]$ intersect, and find the point.

Solution

The directions aren’t multiples. Set the components equal:

$$2 + t = -1 + 2s \quad (x), \qquad 1 + t = 4 - s \quad (y), \qquad t = -1 + s \quad (z)$$

Subtract the $y$ equation from the $x$ equation: $1 = -5 + 3s$, so $s = 2$. Then $t = 1$ from the $y$ equation.

Check $z$: $t = 1$ and $-1 + s = 1$ ✓. The lines intersect at $t = 1$ on $L_1$: $(3, 2, 1)$.

Check on $L_2$ with $s = 2$: $(-1 + 4,\ 4 - 2,\ -1 + 2) = (3, 2, 1)$ ✓.

5. (Core) Classify the lines $L_1: \vec{r} = [1, 1, 0] + t[1, 2, -1]$ and $L_2: \vec{r} = [2, 0, 3] + s[1, 1, 1]$.

Solution

The directions $[1, 2, -1]$ and $[1, 1, 1]$ aren’t multiples, so the lines aren’t parallel.

$$1 + t = 2 + s \quad (x), \qquad 1 + 2t = s \quad (y), \qquad -t = 3 + s \quad (z)$$

Substitute the $y$ equation into the $x$ equation: $1 + t = 3 + 2t$, so $t = -2$ and $s = -3$.

Check $z$: $-t = 2$ but $3 + s = 0$. They don’t match, so the lines are skew.

6. (Core) Find the intersection of the line $\vec{r} = [0, 0, 5] + u[1, 2, -1]$ and the plane $\vec{r} = [1, 0, 0] + s[1, 1, 0] + t[0, 1, 1]$.

Solution

First convert the plane to scalar form. Normal:

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

So $x - y + z + D = 0$, and $(1, 0, 0)$ gives $D = -1$: $x - y + z - 1 = 0$.

Substitute the line, $x = u$, $y = 2u$, $z = 5 - u$:

$$u - 2u + (5 - u) - 1 = 0 \quad\Rightarrow\quad 4 - 2u = 0 \quad\Rightarrow\quad u = 2$$

The point is $(2, 4, 3)$. Check: $2 - 4 + 3 - 1 = 0$ ✓.

7. (Core) For what value of $k$ is the line $\vec{r} = [1, -1, 2] + t[k, 1, 2]$ parallel to the plane $3x - y + 2z = 5$? For that $k$, does the line lie in the plane?

Solution

Parallel means $\vec{m} \cdot \vec{n} = 0$:

$$[k, 1, 2] \cdot [3, -1, 2] = 3k - 1 + 4 = 3k + 3 = 0 \quad\Rightarrow\quad k = -1$$

Test the point $(1, -1, 2)$ in the plane: $3 + 1 + 4 = 8 \ne 5$. So the line is parallel to the plane but not in it.

8. (Challenge) Find the foot of the perpendicular from $P(1, 5, -2)$ to the plane $x + 2y - 2z - 6 = 0$, and the distance from $P$ to the plane.

Solution

The perpendicular line through $P$ has the plane’s normal as its direction: $\vec{r} = [1, 5, -2] + t[1, 2, -2]$. Substitute:

$$\begin{aligned} (1 + t) + 2(5 + 2t) - 2(-2 - 2t) - 6 &= 0 \\ 1 + t + 10 + 4t + 4 + 4t - 6 &= 0 \\ 9t + 9 &= 0 \\ t &= -1 \end{aligned}$$

The foot is $(0, 3, 0)$. Check: $0 + 6 - 0 - 6 = 0$ ✓.

Distance: $\sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{9} = 3$.

9. (Challenge) Find the value of $k$ so that the lines $L_1: \vec{r} = [1, 2, k] + t[1, 0, 1]$ and $L_2: \vec{r} = [3, -1, 0] + s[0, 1, 2]$ intersect. Find the point of intersection.

Solution

Set the components equal:

$$1 + t = 3 \quad (x), \qquad 2 = -1 + s \quad (y), \qquad k + t = 2s \quad (z)$$

The first two give $t = 2$ and $s = 3$. For the lines to meet, the $z$ equation must also hold: $k + 2 = 6$, so $k = 4$.

The point is $(1 + 2,\ 2,\ 4 + 2) = (3, 2, 6)$. Check on $L_2$: $(3,\ -1 + 3,\ 0 + 6) = (3, 2, 6)$ ✓.