Angle between two planes: Worked examples — Maths deep dive

This page is the "throw everything at it" companion to Angle between two planes. We march through every kind of input the topic can give you — clean positive numbers, negative dot products, zeros, parallel planes, vector form, a real-world roof problem, and an exam twist that hides an unknown. For each one you forecast first, then we grind the steps.

Quick reminder of the vocabulary, so no symbol is unearned:

Definition Symbols used on this page

$n = ⟨ a, b, c ⟩$ — the normal vector, the arrow sticking straight out of the plane; its numbers are the coefficients of $x, y, z$ in $a x + b y + cz + d = 0$ . From Equation of a plane.
$r = ⟨ x, y, z ⟩$ — the position vector, the arrow from the origin to a general point $(x, y, z)$ on the plane. Writing $r \cdot n = d$ is just another way of writing $a x + b y + cz = d$ .
$n_{1} \cdot n_{2} = a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2}$ — the dot product, "multiply matching numbers, add them up".
$∣ n ∣ = a^{2} + b^{2} + c^{2}$ — the length (magnitude) of the arrow.
$∣ \cdot ∣$ around the dot product means absolute value: throw away any minus sign, so we always report the acute (small) angle $0^{\circ} \leq θ \leq 9 0^{\circ}$ .

The scenario matrix

Every problem the topic can throw is one of these cells. The last column names the example that covers it.

#	Case class	What's tricky	Covered by
C1	Clean positive dot product	Nothing — warm-up	Ex 1
C2	Negative dot product	Modulus flips it to acute	Ex 2
C3	Zero dot product	Perpendicular planes, $θ = 9 0^{\circ}$	Ex 3
C4	Parallel planes (normals proportional)	$θ = 0^{\circ}$ , not "undefined"	Ex 4
C5	Vector form $r \cdot n = d$	Read normal off directly	Ex 5
C6	Coefficient with a zero ( $c = 0$ , a vertical plane)	Missing variable = normal component $0$	Ex 6
C7	Real-world word problem (two roof panels)	Translate geometry → planes → angle	Ex 7
C8	Exam twist: unknown coefficient, solve for it	Perpendicular condition becomes an equation	Ex 8

Ex 1 — C1: clean positive dot product

Worked example Statement

Find the acute angle between $x + 2 y + 2 z = 9$ and $2 x + 2 y + z = 6$ .

Forecast: The normals $⟨ 1, 2, 2 ⟩$ and $⟨ 2, 2, 1 ⟩$ look similar — both point mostly the same way. Guess: a smallish angle, well under $9 0^{\circ}$ .

Read normals: $n_{1} = ⟨ 1, 2, 2 ⟩$ , $n_{2} = ⟨ 2, 2, 1 ⟩$ . Why this step? Coefficients of $x, y, z$ ARE the normal (from Equation of a plane).
Dot product: $1 \cdot 2 + 2 \cdot 2 + 2 \cdot 1 = 2 + 4 + 2 = 8$ . Why this step? This is the numerator $n_{1} \cdot n_{2}$ ; it measures how aligned the arrows are.
Lengths: $∣ n_{1} ∣ = 1 + 4 + 4 = 3$ , $∣ n_{2} ∣ = 4 + 4 + 1 = 3$ . Why this step? Dividing by lengths strips out "how long" the arrows are, leaving only direction.
Cosine: $cos θ = \frac{∣8∣}{3 \cdot 3} = \frac{8}{9}$ , so $θ = cos^{- 1} \frac{8}{9} \approx 27. 3^{\circ}$ . Why this step? $cos^{- 1}$ answers "which angle has this cosine?"

Verify: $\frac{8}{9} \approx 0.889$ is close to $1$ → the arrows are nearly aligned → small angle. Matches the forecast. ✔

Ex 2 — C2: the negative dot product (modulus earns its keep)

Worked example Statement

Find the acute angle between $2 x - y + 2 z = 3$ and $- x - 2 y + 2 z = 7$ .

Forecast: The second normal has two minus signs. Will the dot product come out negative? If so, do we report an obtuse angle? (Trap!)

Normals: $n_{1} = ⟨ 2, - 1, 2 ⟩$ , $n_{2} = ⟨ - 1, - 2, 2 ⟩$ . Why this step? The coefficients of $x, y, z$ in each equation are exactly the components of that plane's normal (from Equation of a plane) — including their signs, which is what makes this case interesting.
Dot: $2 (- 1) + (- 1) (- 2) + 2 (2) = - 2 + 2 + 4 = 4$ . Why this step? Watch the signs carefully; each term is a signed product.
Lengths: $∣ n_{1} ∣ = 4 + 1 + 4 = 3$ , $∣ n_{2} ∣ = 1 + 4 + 4 = 3$ .
Cosine: $cos θ = \frac{∣4∣}{9} = \frac{4}{9} \approx 0.444$ , $θ \approx 63. 6^{\circ}$ .

Now the pointed lesson: flip the second plane by multiplying by $- 1$ : $x + 2 y - 2 z = - 7$ . Same plane! Its normal is now $⟨ 1, 2, - 2 ⟩$ , and the dot becomes $2 + (- 2) + (- 4) = - 4$ — negative.

Read the figure: the solid black line is the plane seen edge-on. The red arrow $n = ⟨ 1, 2, - 2 ⟩$ points one way; the black dashed arrow $- n = ⟨ - 1, - 2, 2 ⟩$ points the exact opposite way. Both are legitimate normals of the same plane, so the raw cosine can come out $+ \frac{4}{9}$ or $- \frac{4}{9}$ depending on which one you happened to read off. The $∣ \cdot ∣$ collapses both $\pm 4$ into $4$ and rescues the acute answer.

Verify: With modulus, both versions give $cos θ = \frac{4}{9} \Rightarrow 63. 6^{\circ}$ . Without it, one version wrongly reports $18 0^{\circ} - 63. 6^{\circ} = 116. 4^{\circ}$ . The modulus removes this ambiguity. ✔

Ex 3 — C3: zero dot product (perpendicular)

Worked example Statement

Are the planes $x + 2 y - z = 4$ and $3 x - y + z = 0$ perpendicular? If so, that's your angle.

Forecast: Perpendicular means $θ = 9 0^{\circ}$ , which means $cos θ = 0$ , which means the dot product must be exactly $0$ . Predict: does it hit zero?

Normals: $n_{1} = ⟨ 1, 2, - 1 ⟩$ , $n_{2} = ⟨ 3, - 1, 1 ⟩$ . Why this step? We read the coefficients of $x, y, z$ straight off each equation, because those coefficients ARE the normal (from Equation of a plane).
Dot: $1 \cdot 3 + 2 \cdot (- 1) + (- 1) \cdot 1 = 3 - 2 - 1 = 0$ . Why this step? A zero dot product means the normals are perpendicular (see Normal vector and dot product).
Conclude: $cos θ = \frac{∣0∣}{∣ n _{1} ∣∣ n _{2} ∣} = 0 \Rightarrow θ = 9 0^{\circ}$ . Why this step? Perpendicular normals ⇔ perpendicular planes — the wedge is a right angle.

Verify: No length computation even needed: the numerator is $0$ , so no matter what the (nonzero) denominator is, the fraction is $0$ and $cos^{- 1} 0 = 9 0^{\circ}$ . ✔

Ex 4 — C4: parallel planes (angle is $0^{\circ}$ , not "undefined")

Worked example Statement

Find the angle between $2 x - 4 y + 6 z = 1$ and $- x + 2 y - 3 z = 5$ .

Forecast: Look at the coefficients: $2, - 4, 6$ versus $- 1, 2, - 3$ . The second is exactly $- \frac{1}{2}$ of the first. That's a red flag for parallel. Predict $θ = 0^{\circ}$ .

Normals: $n_{1} = ⟨ 2, - 4, 6 ⟩$ , $n_{2} = ⟨ - 1, 2, - 3 ⟩$ . Why this step? Each plane's normal is just its coefficient triple (from Equation of a plane); lining them up side by side is what lets us spot the proportionality in the next step.
Proportionality check: $\frac{2}{- 1} = \frac{- 4}{2} = \frac{6}{- 3} = - 2$ . Why this step? If every ratio is the same, one normal is a scalar multiple of the other → the arrows are collinear → the planes are parallel.
Angle: parallel normals give $θ = 0^{\circ}$ . Sanity via the formula: dot $= 2 (- 1) + (- 4) (2) + 6 (- 3) = - 2 - 8 - 18 = - 28$ ; $∣ n_{1} ∣ = 4 + 16 + 36 = 56$ ; $∣ n_{2} ∣ = 1 + 4 + 9 = 14$ . $cos θ = \frac{∣ - 28∣}{56 14} = \frac{28}{784} = \frac{28}{28} = 1 \Rightarrow θ = 0^{\circ}$ . Why this step? $cos θ = 1$ is the numeric fingerprint of parallel.

Verify: $56 \cdot 14 = 784 = 28$ , and $\frac{28}{28} = 1$ . Parallel confirmed — the planes never form a wedge; the "between" angle is $0^{\circ}$ (they're the same tilt, just shifted). ✔

Ex 5 — C5: vector form $r \cdot n = d$

Worked example Statement

Find the angle between $r \cdot (\hat{i} + \hat{j} - \hat{k}) = 4$ and $r \cdot (2 \hat{i} - \hat{j} + \hat{k}) = 1$ .

Recall $r = ⟨ x, y, z ⟩$ is the position vector of a general point on the plane, so $r \cdot n = d$ literally spells out $a x + b y + cz = d$ .

Forecast: In this form the normal is already sitting there as the vector being dotted. No coefficient-hunting. Guess: moderate angle.

Read normals directly: $n_{1} = ⟨ 1, 1, - 1 ⟩$ , $n_{2} = ⟨ 2, - 1, 1 ⟩$ . Why this step? In the form $r \cdot n = d$ , the vector $n$ being dotted with the position vector $r$ IS the normal — no rearranging needed, because $r \cdot n = d$ is just $a x + b y + cz = d$ in disguise.
Dot: $1 \cdot 2 + 1 \cdot (- 1) + (- 1) \cdot 1 = 2 - 1 - 1 = 0$ . Why this step? Numerator of the formula — and it's zero!
Conclude: $cos θ = 0 \Rightarrow θ = 9 0^{\circ}$ . Why this step? A zero numerator forces $cos θ = 0$ , and the only angle in $[0^{\circ}, 9 0^{\circ}]$ with cosine $0$ is $9 0^{\circ}$ — a right angle.

Verify: Zero numerator ⇒ perpendicular, exactly as in C3, but reached from vector form. Same machine, different disguise. ✔

Ex 6 — C6: a missing variable (vertical plane, zero component)

Worked example Statement

Find the angle between $x + y = 2$ and $x - y + z = 0$ .

Forecast: The first plane has no $z$ — that means $z$ can be anything, so the plane is vertical (runs straight up the $z$ -axis). Its normal must lie flat in the $x y$ -plane. Predict: a real, in-between angle.

Normals (fill the gap with $0$ ): $x + y + 0 z = 2$ gives $n_{1} = ⟨ 1, 1, 0 ⟩$ ; and $n_{2} = ⟨ 1, - 1, 1 ⟩$ . Why this step? A missing variable means its coefficient is $0$ — the normal has a zero component, it just doesn't tilt in that direction.
Dot: $1 \cdot 1 + 1 \cdot (- 1) + 0 \cdot 1 = 1 - 1 + 0 = 0$ . Why this step? The zero $z$ -component silently kills the third term.
Conclude: $cos θ = 0 \Rightarrow θ = 9 0^{\circ}$ . Why this step? When the cosine of the angle is $0$ , the angle whose cosine is $0$ (in the range $0^{\circ}$ to $9 0^{\circ}$ ) is exactly $9 0^{\circ}$ — so the planes meet at a right angle.

Read the figure: the shaded sheet is the plane $x + y = 2$ . Because $z$ is absent, the sheet stands perfectly upright and stretches up the $z$ -axis. The red arrow is its normal $n_{1} = ⟨ 1, 1, 0 ⟩$ — notice it lies flat (its $z$ -component is $0$ ), pointing horizontally away from the sheet, never tilting up or down. That flatness is exactly what "missing $z$ " looks like, and it's why the dot with $⟨ 1, - 1, 1 ⟩$ came out $0$ : the tilted plane's $z$ -lean has nothing to pair with.

Verify: The normal $⟨ 1, 1, 0 ⟩$ lies flat; $⟨ 1, - 1, 1 ⟩$ dotted with it gives $0$ , so they're perpendicular. Geometrically the vertical plane and the tilted plane meet at a right angle. A zero coefficient is not a degenerate error — it's just a normal that doesn't lean along that axis. ✔

Ex 7 — C7: real-world word problem (two roof panels)

Worked example Statement

A tent's two roof panels are modelled as planes. Panel A lies in $2 x + y + 2 z = 6$ ; panel B lies in $x - 2 y + 2 z = 4$ . What angle (to the nearest degree) do the two panels make at the ridge?

Forecast: Roof panels usually meet at a fairly open angle (a shallow tent) or a sharp one (steep tent). Read the normals, expect somewhere in the $5 0^{\circ}$ – $7 0^{\circ}$ range.

Model → normals: Panel A: $n_{1} = ⟨ 2, 1, 2 ⟩$ ; Panel B: $n_{2} = ⟨ 1, - 2, 2 ⟩$ . Why this step? Each physical panel is a plane; its normal is the coefficient triple.
Dot: $2 \cdot 1 + 1 \cdot (- 2) + 2 \cdot 2 = 2 - 2 + 4 = 4$ . Why this step? Measures how aligned the two panel normals are.
Lengths: $∣ n_{1} ∣ = 4 + 1 + 4 = 3$ , $∣ n_{2} ∣ = 1 + 4 + 4 = 3$ . Why this step? These lengths sit in the denominator of the formula; dividing by them cancels the "how long" of each normal so only the direction difference is left — otherwise longer arrows would fake a different angle.
Cosine: $cos θ = \frac{∣4∣}{9} = \frac{4}{9} \approx 0.444$ , $θ \approx 63. 6^{\circ} \approx 6 4^{\circ}$ . Why this step? This is the angle between the normals = the wedge angle between the panels.

Read the figure: the two black lines are the roof panels seen edge-on, meeting at the black dot (the ridge). Out of each panel sticks a red normal arrow, and the red arc between them marks $θ \approx 6 4^{\circ}$ . The key insight the picture makes visible: you never measure the cardboard panels directly — you measure the angle between the two red pencils poking out of them, and that IS the panel-to-panel angle.

Verify (sanity + units): angle is between $0^{\circ}$ and $9 0^{\circ}$ — a sensible roof pitch, not folded flat, not standing straight. Units: an angle in degrees, dimensionless ratio inside $cos^{- 1}$ . $\approx 6 4^{\circ}$ is a believable tent. ✔

Ex 8 — C8: the exam twist (solve for an unknown)

Worked example Statement

For what value of $k$ are the planes $2 x - y + k z = 3$ and $x + 2 y - 4 z = 1$ perpendicular?

Forecast: Perpendicular ⇒ dot product $= 0$ . That gives one equation in $k$ . Predict a single value.

Normals: $n_{1} = ⟨ 2, - 1, k ⟩$ , $n_{2} = ⟨ 1, 2, - 4 ⟩$ . Why this step? $k$ rides along as an unknown coefficient — it's still just the $z$ -component of the normal.
Set dot product to zero: $2 \cdot 1 + (- 1) \cdot 2 + k \cdot (- 4) = 0$ . Why this step? Perpendicular planes ⇔ perpendicular normals ⇔ dot $= 0$ . This is the condition turned into an equation.
Simplify: $2 - 2 - 4 k = 0 \Rightarrow - 4 k = 0 \Rightarrow k = 0$ . Why this step? Solve the linear equation for the unknown.

Verify: Plug $k = 0$ : $n_{1} = ⟨ 2, - 1, 0 ⟩$ , dot with $⟨ 1, 2, - 4 ⟩ = 2 - 2 - 0 = 0$ . ✔ So the first plane is vertical ( $z$ absent) yet meets the tilted plane at exactly $9 0^{\circ}$ — the twist quietly landed us back in C6 territory. ✔

Recall Which cell am I in? (self-test)

Given $x + y + z = 1$ and $2 x + 2 y + 2 z = 9$ , which case class and what's the angle? Case + answer ::: C4 parallel (normals $⟨ 1, 1, 1 ⟩$ , $⟨ 2, 2, 2 ⟩$ proportional), so $θ = 0^{\circ}$ .

Connections

Angle between two planes — the parent formula this page drills.
Scalar (dot) product — the engine: $a \cdot b = ∣ a ∣∣ b ∣ cos θ$ .
Normal vector and dot product — why zero dot ⇒ perpendicular.
Equation of a plane — where the normal $⟨ a, b, c ⟩$ is read off.
Angle between a line and a plane — the $sin$ cousin, don't confuse them.
Angle between two lines — same dot-product machine, different objects.
Distance from a point to a plane — another gadget built on the normal.