C. Geometry and Algebra of Vectors · Lines and Planes

Intersection of a Line and a Plane

C3.3 C4.7About 70 minutes

Learning goals

I can describe the three ways a line and a plane can meet.
I can use m · n to predict the configuration.
I can find the point where a line meets a plane by substituting parametric equations into a scalar equation.
I can find the foot of a perpendicular from a point to a plane, and the angle between a line and a plane.

Words to know

foot of the perpendicular
The point where a line through a given point, perpendicular to a plane (or line), meets that plane (or line).

parallel to a plane
A line is parallel to a plane when its direction is perpendicular to the plane's normal (m · n = 0).

perpendicular to a plane
A line is perpendicular to a plane when its direction is a scalar multiple of the normal.

angle between a line and a plane
The angle φ between the line and its shadow on the plane; sin φ = |m · n|/(|m||n|).

Where does the laser hit the wall?

A laser pointer beam is a line. A wall is a plane. Unless the beam runs exactly parallel to the wall, it hits the wall at one spot. Finding that spot is one of the most useful calculations in this unit. Computer graphics uses calculations like this one, in a technique called ray tracing, to work out which surface a ray of light hits first.

Three cases

A line with direction m and a plane with normal n can meet in:

one point, when m · n ≠ 0 (the line is not parallel to the plane);
no points, when m · n = 0 and a point of the line is not on the plane (the line is parallel to the plane);
infinitely many points, when m · n = 0 and a point of the line is on the plane (the line lies in the plane).

The method: substitute

Write the line in parametric form: x = x₀ + ta, y = y₀ + tb, z = z₀ + tc.
Write the plane in scalar form Ax + By + Cz + D = 0 (convert it first if it is given in vector form).
Substitute the parametric expressions into the plane's equation and solve for t.

The result of step 3 tells you the case:

One value of t: one point. Put t back into the line to find it.
A false statement like 0 = 5: no intersection.
A true statement like 0 = 0: the line lies in the plane.

A line and a plane meet in one point, in no point (the line is parallel to the plane, above or below it), or in every point of the line (the line lies in the plane). The dashed part of the first line is below the plane.

A line perpendicular to a plane

A line is perpendicular to a plane when its direction is parallel to the normal (m = kn). To find the foot of the perpendicular from a point A to a plane, write the line through A with direction n and intersect it with the plane. You will use this in the lesson Distances from a Point to a Line and to a Plane.

The angle between a line and a plane

If θ is the angle between m and n, the line makes the angle φ = 90° − θ with the plane. Since cos θ = sin φ,

sin φ = |m · n| / (|m||n|).

The line (direction m) pierces the plane. The angle φ between the line and the plane is the complement of the angle θ between m and the normal n. If m · n = 0, then θ = 90°, φ = 0° and the line is parallel to the plane.

Worked example

Example 1: One point

Find where the line x = 2 + t, y = −1 + 3t, z = 4 − 2t meets the plane 2x − y + z = 3.

Substitute: 2(2 + t) − (−1 + 3t) + (4 − 2t) = 3.
Simplify: 4 + 2t + 1 − 3t + 4 − 2t = 3, so 9 − 3t = 3 and t = 2.
Point: (2 + 2, −1 + 6, 4 − 4) = (4, 5, 0).
Check: 2(4) − 5 + 0 = 3 ✓.

The line meets the plane at (4, 5, 0).

Worked example

Example 2: Parallel, or lying in the plane

Does the line r = [1, 0, 2] + t[2, 1, −1] meet the plane x − y + z = 5? What about the plane x − y + z = 3?

m · n = [2, 1, −1] · [1, −1, 1] = 2 − 1 − 1 = 0, so the line is parallel to both planes (they are parallel to each other).
Substitute into x − y + z = 5: (1 + 2t) − t + (2 − t) = 5, which gives 3 = 5. False: no intersection.
Substitute into x − y + z = 3: the same left side gives 3 = 3. True for every t: the line lies in that plane.

No intersection with x − y + z = 5. The line lies in x − y + z = 3.

Worked example

Example 3: A perpendicular to a plane given in vector form

Find the point where the perpendicular from A(−5, 3, 7) meets the plane r = [0, 0, 2] + s[−1, 1, 3] + t[2, 0, −3].

Normal: [−1, 1, 3] × [2, 0, −3] = [(1)(−3) − (3)(0), (3)(2) − (−1)(−3), (−1)(0) − (1)(2)] = [−3, 3, −2]. Use n = [3, −3, 2].
Scalar form: 3x − 3y + 2z + D = 0 through (0, 0, 2): 4 + D = 0, so D = −4. Plane: 3x − 3y + 2z − 4 = 0.
Perpendicular line from A: x = −5 + 3u, y = 3 − 3u, z = 7 + 2u.
Substitute: 3(−5 + 3u) − 3(3 − 3u) + 2(7 + 2u) − 4 = 0, so −15 + 9u − 9 + 9u + 14 + 4u − 4 = 0, giving 22u − 14 = 0 and u = 7/11.
Foot: (−5 + 21/11, 3 − 21/11, 7 + 14/11) = (−34/11, 12/11, 91/11).

The perpendicular meets the plane at (−34/11, 12/11, 91/11) ≈ (−3.09, 1.09, 8.27). (The distance from A to the plane is |u| × |n| = (7/11)√22 ≈ 2.98; more on that in the distances lesson.)

Worked example

Example 4: Angle between a line and a plane

Find the angle between the line with direction [1, 2, 2] and the plane 2x − y + 2z = 7.

m · n = 2 − 2 + 4 = 4. |m| = 3 and |n| = 3.
sin φ = 4/9 ≈ 0.444.
φ ≈ 26.4°.

About 26.4°.

Just for fun

A line meeting a plane is like an arrow and a target: it hits exactly once, flies parallel and misses forever, or lies flat on the target, which is technically touching it everywhere but scores no points.

Hands-on activity

Pencil and cardboard

You need: a piece of thin cardboard (a cereal box panel), two pencils, a pushpin and a protractor. First poke a small hole near the middle of the cardboard with the pushpin (point it away from your hand), then widen it with a pencil tip.

Hold the cardboard flat. Stand one pencil straight up on it: this is the normal n.
Hold the second pencil (the line) so it pierces the cardboard through the hole at a slant. It meets the plane in one point. Notice the pencil is not perpendicular to the normal pencil.
Pull it out and hold it above the cardboard, parallel to the surface. Now it is perpendicular to the normal pencil (m · n = 0) and meets the plane nowhere.
Lay it flat on the cardboard. It is still perpendicular to the normal, but now every point is shared.
Estimate, with a protractor, the angle between the slanted pencil and the cardboard, and the angle between the slanted pencil and the normal pencil. They should add to about 90°.

What to look for: m · n = 0 does not decide between "no point" and "every point". You must also check whether a point of the line is on the plane.

Your turn

Find the intersection of x = 1 + t, y = 2 − t, z = 3t with the plane x + y + z = 6.

Describe the intersection of r = [0, 1, 1] + t[1, 1, 0] with x − y + 4z = 3.

Describe the intersection of the same line with x − y + 4z = 7.

Where does r = [1, 2, 3] + t[1, −1, 2] meet the xz-plane?

Find where r = [2, −1, 0] + t[1, 1, 1] meets 3x − y + 2z = 1.

Is the line r = [1, 1, 1] + t[2, −1, 3] perpendicular to the plane 4x − 2y + 6z = 0?

Find the angle between a line with direction [1, 0, 1] and the plane z = 0.

More practice: warm-up

If m · n ≠ 0, how many points do the line and plane share?

If m · n = 0 and a point of the line is on the plane, what is the configuration?

Where does the z-axis meet the plane x + y + z = 5?

Is a line with direction [1, 1, 1] parallel to the plane x − y = 0?

Where does x = t, y = t, z = t meet x + y + z = 3?

More practice: on your own

Find where the line (x − 1)/2 = (y + 1)/(−1) = (z − 2)/3 meets x + 2y − z = 4.

Find k so that r = [1, 0, 0] + t[2, k, 1] is parallel to x + 3y − 5z = 2. Does the line lie in the plane?

Find the foot of the perpendicular from P(1, 2, −1) to the plane 2x + y − z = 9.

Where does the line through A(1, 1, 1) and B(3, 5, −1) meet the plane x − y + z = 0?

Find the angle between a line with direction [2, 1, 2] and the plane x + y = 3.

Challenge

Find where the line r = [2, 3, 1] + u[1, 2, 3] meets the plane r = [1, 0, 0] + s[1, 1, 0] + t[0, 1, 1].

Find the reflection of P(1, 2, 3) in the plane x + y + z = 0.

Find a and b so that the line r = [1, a, 2] + t[1, 1, b] lies entirely in the plane 2x − y + z = 3.

Check your understanding

Quiz

Where does the line x = t, y = 2t, z = 3t meet the plane x + y + z = 12?

A line has m · n = 0, and a point on the line does not satisfy the plane's equation. The line and plane:

Describe the intersection of r = [0, 1, 1] + t[1, 1, 0] and x − y + 4z = 3.

Which is a direction vector of a line perpendicular to the plane 2x − y + 3z = 5?

The angle φ between a line (direction m) and a plane (normal n) satisfies:

Where does r = [1, 2, 3] + t[1, −1, 2] meet the plane y = 0?

For parents and teachers

Big idea (C3.3, C4.7): a line meets a plane in one point, no point or every point. Substituting the line's parametric equations into the plane's scalar equation gives one equation in t, and its result (one value, a false statement or a true statement) tells you which case you have.

How to help: Ask your teen to predict the case first using m · n, then substitute. Always check the final point in the plane's equation.

Common difficulties:

Trying to substitute into a plane given in vector form. Convert it to scalar form first.
Seeing 0 = 0 and writing "no solution". A true statement means the line lies in the plane.
Using cos instead of sin for the line-plane angle (that gives the angle with the normal).
Forgetting to put t back into the line to get the point.