Hand a line its direction and a plane its normal, and parallel and perpendicular become arithmetic.
A line in space runs on forever, and a plane spreads out without edge — yet each one can be pinned down by a single arrow. Point an arrow the way a line travels and that one direction vector d stands in for the whole line. Stand an arrow straight up out of a plane, like a flagpole on a field, and that one normal vector n stands in for the whole surface. Once a line is a vector and a plane is a vector, the two big questions of solid geometry — are these parallel? and are these perpendicular? — stop being puzzles about hidden lines and become two pieces of arithmetic: is one vector a multiple of the other? and is the dot product zero? This lesson reuses everything from Stage 27 — coordinates, the scalar multiple b = λa, and the dot product a·b = x₁x₂ + y₁y₂ + z₁z₂ — and turns it into a perpendicularity detector.
A line has a place and a heading. Fix any point on the line, then choose any arrow d that points along the line, and you have caught it completely: the line is "all points you reach from that point by sliding some multiple of d." We call d a direction vector of the line. There is nothing special about its length — d, 2d, and −d all describe the very same line, just as a road sign points the same way whether the arrow is long or short.
The easiest direction vector to write down comes from two points on the line. If the line passes through A and B, then AB→ = B − A runs straight along it. For example, a line through A(1, 0, 2) and B(3, 1, 2) has direction AB→ = (3−1, 1−0, 2−2) = (2, 1, 0) — computed, not guessed.
A plane needs a different kind of arrow. Pointing along a plane doesn't capture it — a plane has infinitely many directions inside it. What is unique is the direction away from it. The arrow that stands perpendicular to a plane — square to every line drawn inside it — is the plane's normal vector n. Think of a flat field: countless paths cross it, but exactly one direction is "straight up." That up-arrow is the field's normal, and it tells you the plane's tilt unambiguously.
Like a direction vector, a normal isn't unique in length or sign: n, 3n, and −n are all normals of the same plane. What pins the plane down is the line the normal runs along. Two planes are parallel exactly when their normals point the same way; two planes meet at a tilt set by the angle between their normals.
A direction vector d represents a line (it lies along it). A normal vector n represents a plane (it stands across it). One points with the road; the other points out of the ground.
How do you actually get a normal? A plane is laid out by two non-parallel arrows u and v sitting inside it. The normal n must be perpendicular to both, so it must satisfy two equations at once:
n·u = 0 and n·v = 0.
One clean route solves those equations directly (set one component to 1 and solve for the rest). The faster route uses the cross product from 29.2: n = u × v is built to be perpendicular to both u and v, so it is a normal automatically.
A tilted plane is spanned by u = (1, 0, 1) and v = (0, 1, 1). Then
n = u × v = (0·1 − 1·1, 1·0 − 1·1, 1·1 − 0·0) = (−1, −1, 1).
Check it: n·u = (−1)(1) + (−1)(0) + (1)(1) = 0 ✓ and n·v = (−1)(0) + (−1)(1) + (1)(1) = 0 ✓. Both zero — (−1, −1, 1) truly stands square to the plane.
If you are handed three points P, Q, R instead of two arrows, just build the arrows first: take u = Q − P and v = R − P, then cross them. As long as the three points aren't in a straight line, u and v aren't parallel and the cross product is a genuine normal.
A normal must be perpendicular to two non-parallel in-plane vectors — not just one. An arrow that is ⊥ to a single line in the plane can still lean along the surface; only being ⊥ to two independent directions forces it straight out. (That is exactly why u × v needs two different arrows: if u ∥ v, the cross product collapses to the zero vector and gives no normal at all.)
Now the payoff. Every "is this parallel to that?" question becomes a vector check.
That last one trips up nearly everyone, so let's see why it has to be true. If a line runs parallel to a plane, it never climbs toward or away from the surface — so its direction lies flat in the plane's tilt, which is exactly the directions that are perpendicular to the up-arrow n. A line parallel to a plane is "going nowhere in the normal's direction," and "going nowhere along n" is precisely d·n = 0.
Set axes at corner A of a cube of side 2, so its base ABCD sits at z = 0 with normal n = (0, 0, 1). The top diagonal runs A₁(0,0,2) to C₁(2,2,2), giving direction d = A₁C₁→ = (2, 2, 0).
d·n = (2)(0) + (2)(0) + (0)(1) = 0,
and since the diagonal sits up at z = 2, it is not in the base. So the top diagonal is parallel to the base plane — proven by one multiplication and an addition, no auxiliary lines drawn.
The perpendicular tests are just as short — but notice they flip "multiple" and "dot = 0" compared with the parallel tests.
The vertical edge A(0,0,0) to A₁(0,0,2) has direction d = AA₁→ = (0, 0, 2). The base normal is n = (0, 0, 1). Then d = 2n, so d ∥ n — the edge is perpendicular to the base, exactly as a cube's upright edge should be.
And the two base-edges AB→ = (2, 0, 0) and AD→ = (0, 2, 0) dot to (2)(0) + (0)(2) + (0)(0) = 0: AB ⊥ AD, the square corner of the floor.
A line parallel to a plane does NOT mean d ∥ n. Parallel-to-a-plane means the direction is perpendicular to the normal: d ⊥ n, i.e. d·n = 0. It is d ∥ n (direction is a multiple of the normal) that makes a line perpendicular to the plane. The two cases are mirror images — keep them straight: d·n = 0 ⇒ line ∥ plane; d = λn ⇒ line ⊥ plane.
Here is the whole method in one sentence: set up coordinates, write down the direction vectors and normal vectors, and read off ∥ or ⊥ from a scalar-multiple test or a single dot product. No clever auxiliary line, no chasing equal angles around a figure — just the same two checks over and over. The widget below lets you fire those checks on a cube: pick a line (an edge or a diagonal, which gives d) and a face (which gives n), and the verdict pops out of d·n.
Spin through the combinations and a pattern emerges: the upright edge AA₁ is ⊥ to the base (its d ∥ n), parallel to the opposite side face (its d ⊥ that normal), and actually lies inside the side face it touches — the widget separates that last case out, since "d·n = 0" only proves the line doesn't climb away from the plane, not that it floats above it. The top diagonal A₁C₁ rides truly ∥ to the base; the space diagonal AC₁ is neither ∥ nor ⊥ to any face — it stabs through at a slant. Every one of those verdicts came from a single dot product (plus, for "∥," one check that a point of the line is off the plane).
A line carries a direction vector d; a plane carries a normal vector n = u × v. Then every parallel/perpendicular question is one of two checks — "is it a multiple?" or "is the dot product zero?" Watch how ∥ and ⊥ swap places between the line–plane row and the others.
| Objects | Parallel ∥ | Perpendicular ⊥ |
|---|---|---|
| line & line | d1 = λ d2 (multiple) | d1·d2 = 0 |
| line & plane | d·n = 0 (and not in it) | d = λ n (multiple) |
| plane & plane | n1 = λ n2 (multiple) | n1·n2 = 0 |
To find a normal from two in-plane arrows: n = u × v, then confirm n·u = 0 and n·v = 0. The trap to memorize: d·n = 0 means line ∥ plane, while d ∥ n means line ⊥ plane.
A line passes through A(2, −1, 0) and B(2, 3, 4). Write down a direction vector for the line.
AB→ = B − A = (2−2, 3−(−1), 4−0) = (0, 4, 4). (Any multiple, such as (0, 1, 1), is equally valid.)
A plane is spanned by u = (2, 0, 0) and v = (0, 1, 3). Find a normal vector and verify it.
n = u × v = (0·3 − 0·1, 0·0 − 2·3, 2·1 − 0·0) = (0, −6, 2). Check: n·u = 0 ✓ and n·v = 0·0 + (−6)(1) + 2·3 = 0 ✓.
A line has direction d = (1, 2, −1) and a plane has normal n = (2, −1, 0). Is the line parallel to the plane, perpendicular to it, or neither?
d·n = (1)(2) + (2)(−1) + (−1)(0) = 2 − 2 + 0 = 0, so d ⊥ n and the line is parallel to the plane (assuming it isn't lying in it). It is not perpendicular — that would need d = λn, and (1, 2, −1) is no multiple of (2, −1, 0).
Two lines have directions d1 = (1, 2, 2) and d2 = (2, −2, 1). Are the lines perpendicular?
d1·d2 = (1)(2) + (2)(−2) + (2)(1) = 2 − 4 + 2 = 0. Yes — the dot product is zero, so the two lines are perpendicular.
A plane passes through P(1, 0, 0), Q(0, 1, 0), R(0, 0, 1). A line has direction d = (1, 1, 1). Show the line is perpendicular to the plane.
Build two in-plane arrows: u = Q − P = (−1, 1, 0), v = R − P = (−1, 0, 1). Then n = u × v = (1, 1, 1). Since d = (1, 1, 1) = 1·n, we have d ∥ n, so the line is perpendicular to the plane.
In a cube ABCD-A₁B₁C₁D₁ of side 2 with axes at A, prove the space diagonal AC₁ is perpendicular to the base diagonal BD.
With axes at A: B(2,0,0), D(0,2,0), C₁(2,2,2). So AC₁→ = (2, 2, 2) and BD→ = D − B = (−2, 2, 0). Their dot product is (2)(−2) + (2)(2) + (2)(0) = −4 + 4 + 0 = 0, so AC₁ ⊥ BD.
Six questions to lock it in. Tap the answer you think is right.
This lesson realizes the heart of HSN-VM (Vector & Matrix Quantities) — representing lines and planes by vectors and using the dot product as a perpendicularity test — and feeds HSG-GMD and G-MG (geometric modeling in space) by converting "is this line parallel to that face?" into a coordinate calculation. The single most common error to head off is the line-versus-plane trap: a line parallel to a plane satisfies d·n = 0 (direction perpendicular to the normal), while a line perpendicular to a plane satisfies d = λn (direction along the normal). Encourage students to always verify a normal with the two checks n·u = 0 and n·v = 0, and to remember that a normal must be perpendicular to two non-parallel directions, not just one.