The Intersection of a Line and a Plane

One point on both

A line in space is r = a + t d. Each value of the parameter t gives one point of the line. A plane is an equation, such as 2x + y + z = 12, that its points satisfy and other points do not.

Where the line meets the plane, there is a point that is on both: a point of the line whose coordinates satisfy the plane’s equation. So the question is which value of t gives such a point.

Take the line r = (1, 0, 2) + t(1, 1, 1) and the plane 2x + y + z = 12.

Substitute the line into the plane

Write the line one coordinate at a time: x = 1 + t, y = t and z = 2 + t.

Put these into the plane’s equation in place of x, y and z: 2(1 + t) + t + (2 + t) = 12. Multiply out the bracket and collect terms: 2 + 2t + t + 2 + t = 12, which is 4t + 4 = 12.

Three unknowns x, y and z have become one, t, so a single equation is enough to find it.

The value of 2x + y + z at the point of the line with parameter t is 4t + 4: it starts at 4 when t = 0 and rises by 4 for each step along the line. It reaches the plane’s 12 once, at t = 2.

Solve, then put t back

Take 4 off both sides: 4t = 8. Divide by 4: t = 2.

t = 2 is not the answer yet; it only says how far along the line the meeting point is. Put it back into the line: r = (1, 0, 2) + 2(1, 1, 1) = (1, 0, 2) + (2, 2, 2) = (3, 2, 4).

Check by substituting the point into the plane: 2 × 3 + 2 + 4 = 6 + 2 + 4 = 12. The point (3, 2, 4) is on the line and on the plane.

A meeting behind the start

Take the line r = (2, 1, 0) + t(1, 2, 3) and the plane 2x + y + z = −2. Then x = 2 + t, y = 1 + 2t and z = 3t, and substituting gives 2(2 + t) + (1 + 2t) + 3t = −2, which is 5 + 7t = −2.

So 7t = −7 and t = −1. A negative t is fine: the meeting point is one step behind the starting point, at (2, 1, 0) − (1, 2, 3) = (1, −1, −3). Check: 2 × 1 + (−1) + (−3) = 2 − 1 − 3 = −2.

When the t terms cancel

Substituting r = a + t d into a plane r · n = k always gives a · n + t(d · n) = k. The number in front of t is d · n, the dot product of the line’s direction with the plane’s normal. In the first example, (1, 1, 1) · (2, 1, 1) = 4, which is the 4 in 4t + 4.

If d · n is not 0, there is exactly one t, and the line crosses the plane at one point. If d · n = 0, the direction is perpendicular to the normal, so the line runs parallel to the plane and the t terms cancel. What is left is a statement with no t in it, and it is either false for every t or true for every t.

The line r = (1, 0, 2) + t(1, −1, −1) has d · n = 2 − 1 − 1 = 0 with the plane 2x + y + z = 12. Substituting gives 2(1 + t) + (−t) + (2 − t) = 12, which is 4 + 0t = 12, or 0t = 8. No t makes 0 equal 8, so the line never meets the plane: it runs parallel to it and clear of it.

The line r = (3, 2, 4) + t(1, −1, −1) has the same direction and starts at a point of the plane. Substituting gives 2(3 + t) + (2 − t) + (4 − t) = 12, which is 12 + 0t = 12, or 0t = 0. Every t works, so every point of the line is on the plane: the line lies in it.

Along the line r = (1, 0, 2) + t(1, −1, −1), the value of 2x + y + z stays at 4 for every t, because the t terms cancel. It never reaches the plane’s 12, so the line never meets the plane. Started at (3, 2, 4) instead, the value would sit at 12 all the way along.

The same three outcomes, one dimension down: two lines in a plane, y = x + 1 and a second line whose gradient $m_2$m₂ and intercept $c_2$c₂ are the handles. Different gradients give one crossing point. Make the gradients equal with a different intercept and the lines never meet; make the intercepts equal too and they share every point.

The usual mistakes

Dividing before subtracting. From 4t + 4 = 12, dividing 12 by 4 gives 3, which is wrong; take the 4 off first, then divide 8 by 4.

Stopping at 4t = 8. That is four times t; t itself is 2.

Giving t as the meeting point, or walking one stride instead of t strides. The point is a + t d with the t you found: two strides of (1, 1, 1) from (1, 0, 2).

Reading 0t = 8 as one meeting point. There is no t to solve for once the t terms have canceled; a false statement means no meeting at all, and a true one, such as 0t = 0, means the line lies in the plane.

A drone landing on a hillside

In the application below, the parameter is time: $\lambda$λ seconds after it starts, a drone is at a point of a straight line, and the hillside is a plane. Substituting the drone’s coordinates into the hillside’s equation gives the moment it lands, and putting that moment back gives where.