Analytic geometry · Space
A line and a plane in space can relate in only three ways: the line pierces the plane in one point, runs parallel to it, or lies entirely in it. Which case occurs, and at what angle the line meets the plane, is decided by two vectors: the line's direction \(\vec p\) and the plane's normal \(\vec N\). This post works through the three cases and then uses them to solve the classic problems: projecting a point, reflecting it, and projecting a whole line onto a plane.
Throughout, the line and the plane are
\[ p:\ \frac{x - x_1}{l} = \frac{y - y_1}{m} = \frac{z - z_1}{n}, \quad \vec p = \{l, m, n\}; \qquad \alpha:\ Ax + By + Cz + D = 0, \quad \vec N = \{A, B, C\}. \]The line passes through \(M_1(x_1, y_1, z_1)\) in direction \(\vec p\); the plane is perpendicular to \(\vec N\). Our running example will be
\[ p:\ \frac{x - 1}{2} = \frac{y + 1}{1} = \frac{z - 2}{-1}, \qquad \alpha:\ x + 2y + 2z - 6 = 0. \]The angle \(\theta\) between a line and a plane is the angle between the line and its orthogonal projection onto the plane, so \(0^\circ \le \theta \le 90^\circ\). The normal \(\vec N\) stands at \(90^\circ\) to the plane, so the angle between \(\vec p\) and \(\vec N\) is \(90^\circ - \theta\). Its cosine is therefore \(\sin\theta\):
Why this is so convenient: the parametric form describes every point of the line with a single number \(t\). The question "which point of the line is in the plane?" becomes one linear equation in one unknown:
\[ (Al + Bm + Cn)\,t + (Ax_1 + By_1 + Cz_1 + D) = 0. \tag{*} \]Equation (*) has the form \( a\,t + b = 0 \) with \( a = Al + Bm + Cn = \vec N\cdot\vec p \) and \( b = Ax_1 + By_1 + Cz_1 + D \) (the plane's equation evaluated at \(M_1\)). A linear equation has one solution, none, or infinitely many, and these are exactly the three geometric cases:
| Condition | Equation (*) | Position |
|---|---|---|
| \( Al + Bm + Cn \ne 0 \) | exactly one \(t\) | the line pierces the plane at one point |
| \( Al + Bm + Cn = 0 \), \( Ax_1 + By_1 + Cz_1 + D \ne 0 \) | no solution | the line is parallel to the plane, outside it |
| \( Al + Bm + Cn = 0 \), \( Ax_1 + By_1 + Cz_1 + D = 0 \) | every \(t\) is a solution | the line lies in the plane |
The two conditions are geometric: one point of the line is in the plane, and the line runs parallel to the plane. Together they force the whole line into the plane. Either one alone is not enough: the first only says the line meets the plane at \(M_1\), the second only that it is parallel.
The tools above (parametric form, \(\vec p\), \(\vec N\), cross product) solve a family of standard problems. Each needs only a short idea.
Its direction is the normal: through \(A(a_1, a_2, a_3)\), the line is \( \frac{x - a_1}{A} = \frac{y - a_2}{B} = \frac{z - a_3}{C} \).
Its normal is the direction: \( l(x - a_1) + m(y - a_2) + n(z - a_3) = 0 \).
The plane contains the direction \(\vec p\) and the vector \( \overrightarrow{M_1B} \) from a point of the line to the given point \(B\), so its normal is \( \vec p\times\overrightarrow{M_1B} \).
The projection \(p'\) of a line \(p\) onto \(\alpha\) is the intersection of \(\alpha\) with the projecting plane: the plane through \(p\) perpendicular to \(\alpha\). That plane contains \(\vec p\) and \(\vec N\), so its normal is \( \vec p\times\vec N \).
| Question | Answer |
|---|---|
| Angle | \( \sin\theta = \frac{|Al + Bm + Cn|}{|\vec N||\vec p|} \) |
| Parallel | \( Al + Bm + Cn = 0 \) |
| Perpendicular | \( \frac Al = \frac Bm = \frac Cn \) |
| Intersection point | substitute the parametric form, solve for \(t\) |
| Line lies in the plane | \( Ax_1 + By_1 + Cz_1 + D = 0 \) and \( Al + Bm + Cn = 0 \) |
| Projection of a point | intersect the perpendicular (direction \(\vec N\)) with the plane |
| Projection of a line | intersect \(\alpha\) with the plane through \(p\) with normal \(\vec p\times\vec N\) |
Try each one before opening the solution. More solved problems from this topic are in Geometry.
\( \vec p = \{1, 1, 0\} \), \( \vec N = \{0, 0, 1\} \): \( \sin\theta = 0 \), \(\theta = 0\). The line lies in the plane \(z = 0\) (it passes through the origin).
\( (2 - t) + (1 + t) + 3t = 9 \Rightarrow 3 + 3t = 9 \Rightarrow t = 2 \): the point \((0, 3, 6)\).
\( 3k - 2 + 4 = 0 \Rightarrow k = -\frac23 \). The point \((1, 0, -1)\): \( 3 - 0 - 4 = -1 \ne 0 \), so the line is parallel to the plane but not in it.
\( (1 + t, t, t) \): \( 1 + 3t = 4 \), \( t = 1 \). Foot \((2, 1, 1)\); distance \( |t|\cdot|\vec N| = \sqrt3 \).
\( \frac{x - 1}{2} = \frac{y - 2}{-1} = \frac{z - 3}{5} \). Substituting \( (1 + 2t, 2 - t, 3 + 5t) \): \( 2 + 4t - 2 + t + 15 + 25t = 0 \), \( t = -\frac12 \): the point \( \big(0, \frac52, \frac12\big) \).
Direction \(\{1, -1, 0\}\): \( 1 - 1 + 0 = 0 \) (parallel). Point \((0, 1, 2)\): \( 0 + 1 + 10 - 11 = 0 \) (on the plane). Both conditions of 3° hold.
All questions about a line and a plane come down to two numbers: \( \vec N\cdot\vec p = Al + Bm + Cn \), which compares the line's direction with the plane's normal, and \( Ax_1 + By_1 + Cz_1 + D \), which says whether a point of the line is in the plane. The first gives the angle (through a sine, since the normal is at \(90^\circ\) to the plane) and decides between piercing and parallel; the second separates "parallel outside" from "lying in". Substituting the parametric form into the plane's equation captures all of this in one line of algebra, and the projection and reflection problems are short applications of the same idea.
Related posts: the plane in space and the line in space.