Drop a shadow on the normal and read the distance straight off the coordinates.
Here is the prize at the end of the strand. A "how far is it?" question in space — how far is this corner from that slanted face? how thick is the gap between two parallel shelves? — has, until now, meant dropping auxiliary perpendiculars and chasing right triangles through a tangle of edges. We are going to make all of that vanish. The trick is the same one that powered every earlier lesson: a distance is the length of a shadow, and a shadow is a dot product. Set coordinates at a square corner, write down a normal n for the plane, project the arrow to your point onto that normal, and the distance falls out as one quotient: d = |QP→·n| ⁄ |n|. Lines and parallel planes reduce to the same formula, and a five-step routine ties parallelism, perpendicularity, angles, and distances into a single calculation — the doorway to Stage 30, Analytic Geometry.
Everything below stands on the dot product from Stage 27 — its geometric meaning is exactly the shadow we are about to cast — and on the spatial machinery of 29.2 (coordinates), 29.3 (a plane's normal n), and 29.4 (every angle as one cosine). If those tools feel shaky, glance back; if not, read on.
Shine a light straight down onto a line and an arrow casts a shadow on it. That shadow is the projection, and it is the entire geometric content of the dot product. Take a vector a and a direction b. Drop a perpendicular from the tip of a onto b's line; the segment from the shared tail to the foot is a's shadow on b.
Its signed length — the scalar projection — is
scalar projection of a onto b = a·b|b| .
Why? Because a·b = |a||b|cos θ, so dividing by |b| leaves |a|cos θ — exactly the leg of the right triangle that is the shadow. The sign carries information: it is positive when the shadow points the same way as b (θ acute), zero when a ⊥ b (the shadow collapses to a point), and negative when the shadow points backward (θ obtuse). The vector projection just reattaches the direction: multiply the scalar by the unit vector b̂ = b⁄|b|.
Let a = (2, 2, 1) and b = (3, 0, 0). Then a·b = 2·3 + 2·0 + 1·0 = 6, and |b| = √(9+0+0) = 3. So the scalar projection is 63 = 2, and the vector projection is 2·b̂ = 2·(1, 0, 0) = (2, 0, 0) — a lies 2 units "along" the x-direction, no matter how tall it stands. The right angle at the foot is the whole point: the drop is perpendicular, so the shadow is the closest b-multiple to a.
Now point the light along a plane's normal n. Here is the key picture: a point P sits some distance above a plane α, and the distance we want is the length of the perpendicular drop from P to the plane. That drop runs in the n-direction (the normal is, by definition, perpendicular to the surface). So the distance is just the shadow of a connecting arrow on n.
Pick any point Q that lies in the plane and form the arrow QP→ = P − Q. Its shadow on the unit normal n̂ has length |QP→·n̂|, which is the perpendicular drop. Writing n̂ = n⁄|n|:
d = |QP→·n||n| .
The absolute value is there because we want a length, not a signed one — P may sit on either side of the plane. And the answer does not depend on which in-plane point Q you choose: move Q to a different spot Q′ and the arrow changes by QQ′→, which lies in the plane and is therefore ⊥ n — its shadow on n is zero, so the quotient is unchanged. That freedom is what makes the formula so easy to use: grab the most convenient corner.
Plane α: x + 2y + 2z = 1, so a normal is n = (1, 2, 2) with |n| = √(1+4+4) = 3. Take Q = (1, 0, 0) on it and the target P = (2, 2, 2). Then QP→ = P − Q = (1, 2, 2), and QP→·n = 1·1 + 2·2 + 2·2 = 9. So d = |9|3 = 3. The foot of the drop lands back at (1, 0, 0), squarely on the plane — no auxiliary triangle anywhere.
1. Distance is the shadow on n, not along the plane. The drop is perpendicular to α — projected onto the normal. Projecting QP→ onto a vector lying in the plane gives how far you slide across the surface, which is the opposite of what you want. Shadow on the surface ≠ distance to the surface.
2. Divide by |n|. The formula needs the unit normal. If n isn't a unit vector — and it usually isn't — forgetting the |n| in the denominator inflates the answer by a factor of |n|. Here that would have given 9 instead of 3.
3. Any Q on the plane is fine — but it must be on the plane. The point Q has to actually satisfy the plane's equation; a stray nearby point quietly breaks everything.
The beauty of the point-to-plane formula is how much it swallows.
Line ∥ plane. When a line runs parallel to a plane, every point of the line is exactly the same height above the plane (that is what parallel means). So the line-to-plane distance is just the point-to-plane distance of any one point you pick on the line. One calculation, done.
Plane ∥ plane. Two parallel planes are a fixed slab apart. Grab any single point on the first plane and measure its distance to the second — that is the gap. Plane-to-plane collapses to point-to-plane as well.
Point to a line. The distance from a point P to a line ℓ (direction d, through A) is the perpendicular component of AP→ — drop the part that runs along ℓ and keep what's left. In symbols, subtract the vector projection from the arrow:
d = | AP→ − (the shadow of AP→ on d) | = | AP→ − AP→·d|d|2 d | .
Same idea, mirrored: for a plane you keep the shadow on the normal; for a line you throw away the shadow on the direction and keep the rest. Either way, a distance is a projection.
Every "is it parallel / perpendicular / what is the angle / how far" problem in space now follows one checklist. Memorize the five moves and you never need a clever auxiliary line again.
| Step | What you do | The tool |
|---|---|---|
| 1 | Find a square corner and set the x, y, z axes along its three edges. | a cube/box corner |
| 2 | Write every relevant point's coordinates (x, y, z). | read off the box |
| 3 | Build the direction vectors d (B − A) and normals n (n = u×v) you need. | vsub3, vcross3 |
| 4 | Plug into the right formula — a dot for ⊥, a multiple for ∥, sin/cos for an angle, |QP→·n|/|n| for a distance. | vdot3 |
| 5 | Read the answer — and sanity-check it against the picture. | conclude |
The labor that used to be geometric — finding the right perpendicular, naming the right triangle — is replaced by step 1 alone. After you choose the corner, it is pure arithmetic on triples.
Let's run one problem that asks for all four relationships at once, so you can watch the five steps dispatch them in a single setup. Take a cube of side 2 and set axes at corner A: then
A(0,0,0) B(2,0,0) C(2,2,0) D(0,2,0), and the top A₁(0,0,2) B₁(2,0,2) C₁(2,2,2) D₁(0,2,2).
(a) Parallel? Is the top diagonal A₁C₁ parallel to the base ABCD? Direction d = C₁ − A₁ = (2, 2, 0); the base normal is n = (0, 0, 1). Test d·n = 2·0 + 2·0 + 0·1 = 0. So d ⊥ n, and since A₁C₁ is not in the base, A₁C₁ ∥ base.
(b) Perpendicular? Is the edge AA₁ perpendicular to the base? AA₁ = (0, 0, 2), which is 2·(0, 0, 1) = 2n — a scalar multiple of the base normal. Direction ∥ normal means AA₁ ⊥ base. (A vertical edge stands on the floor; the algebra agrees.)
(c) What angle? The angle between the space diagonal AC₁ = (2, 2, 2) and the base uses the line–plane formula sin θ = |d·n|/(|d||n|). Here d·n = (2, 2, 2)·(0, 0, 1) = 2, |d| = √12 = 2√3, |n| = 1, so sin θ = 22√3 = 1⁄√3 ≈ 0.577, giving θ ≈ 35.3°.
(d) How far? Distance from corner C = (2, 2, 0) to the slanted plane through A, B₁, D₁. Spanners u = B₁−A = (2, 0, 2) and v = D₁−A = (0, 2, 2), so n = u×v = (−4, −4, 4) with |n| = 4√3. Take Q = A; then QC→ = (2, 2, 0) and QC→·n = −8 − 8 + 0 = −16. So d = |−16|4√3 = 4⁄√3 ≈ 2.31. One coordinate setup, four answers.
Notice what we never drew: not one auxiliary perpendicular, not one extra triangle. Step 1 (axes at A) carried the whole burden; steps 2–5 were arithmetic. That is the payoff of vectors in space — geometry becomes calculation.
Step back and see what just happened to geometry. A point became a triple of numbers. A line became a point plus a direction vector. A plane became a point plus a normal — or, equivalently, an equation like x + 2y + 2z = 1, whose coefficients are the normal. Parallel, perpendicular, angle, distance: all of them turned into arithmetic on those numbers.
The natural next move is to run the trade in reverse. If a plane is captured by an equation, why not let an equation capture a circle, a parabola, an ellipse — and then let the equation draw the curve? That is exactly the program of Stage 30 · Analytic Geometry: coordinates become equations, and equations become curves. You have built the bridge; next stage you cross it.
Every distance in space is a projection, and every projection is a dot product. Keep this table within reach.
| What you want | The formula | In words |
|---|---|---|
| Scalar projection of a on b | a·b|b| | the signed length of a's shadow on b |
| Vector projection of a on b | a·b|b|2 b | that shadow, with its direction reattached |
| Point P to plane α | |QP→·n||n| | Q any plane point; the drop onto the normal |
| Point P to line ℓ | |AP→ − projdAP→| | throw away the part along d, keep the rest |
| Line ∥ plane, or plane ∥ plane | = point-to-plane d | measure one point — all points share it |
And the five-step routine: corner → coordinates → directions & normals → formula → answer.
Cast the shadow. Find the scalar and vector projections of a = (1, 2, 2) onto b = (0, 0, 4).
Point to plane. How far is P = (4, 1, 2) from the plane 2x + y + 2z = 4?
Don't forget the denominator. A student computes the distance from the origin to the plane x + 2y + 2z = 6 and writes d = 6. What did they forget, and what is the right answer?
In the cube. Cube of side 2, axes at A: A(0,0,0), B(2,0,0), C₁(2,2,2), D(0,2,0). Find the distance from B to the plane through A, C₁, and D.
Point to line. Find the distance from P = (2, 2, 2) to the x-axis (the line through O with direction d = (1, 0, 0)).
A parallel gap. In the same cube of side 2, show that the top diagonal A₁C₁ is parallel to the base ABCD, and find the distance between them.
Six questions to lock it in. Tap the answer you think is right.
This lesson is the modeling payoff of the strand: it turns a hard solid-geometry distance into a single dot-product calculation. The vector machinery — projection, the unit normal, the point-to-plane formula d = |QP→·n|/|n| — lives squarely in HSN-VM (Vector & Matrix Quantities), building on dot products and their geometric meaning. Using coordinates to model and measure a physical solid — choosing axes at a corner, finding heights, gaps, and angles inside a cube — is the geometric-modeling work of HSG-GMD and G-MG (Modeling with Geometry). A good prompt at home: ask your student to name the five steps out loud before touching any numbers — the discipline of "corner, coordinates, directions and normals, formula, answer" is more durable than any single computation, and it is exactly the habit of mind Stage 30 (Analytic Geometry) will reward.