Distance formula in 3D: Worked examples — Maths deep dive

Everything here rests on the one rule from the parent note: $P Q = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} + (z_{2} - z_{1})^{2}$ Read that as "same minus same, square, add, root." Nothing new is invented below — we only stress-test it.

The scenario matrix

Before solving anything, let's list every kind of situation this topic can throw at you. Each row is a "cell"; every worked example below is tagged with the cell it covers.

Cell	Situation	What makes it tricky	Covered by
A	All-positive coordinates, plain plug-in	nothing — the warm-up	Ex 1
B	Mixed signs (negatives)	subtracting a negative flips to a plus	Ex 2
C	A point on an axis / plane (some coords zero)	a whole term vanishes	Ex 3
D	Degenerate: the two points coincide	distance must be exactly $0$	Ex 4
E	Solve for an unknown coordinate	one equation, possibly two answers	Ex 5
F	Geometry proof (type of triangle)	compare three distances	Ex 6
G	Real-world word problem (ant in a room)	translate words → coordinates	Ex 7
H	Exam twist: equidistant point / locus	set two distances equal, cancel squares	Ex 8

Once all eight cells are green, you have seen the full behaviour of the formula.

Cell A — plain plug-in

Cell B — negative coordinates

Negatives are where most marks are lost, so go slowly.

Worked example Example 2 — mixed signs

Find the distance between $A (- 1, 2, - 3)$ and $B (2, - 2, - 15)$ .

Forecast: With $z$ going from $- 3$ to $- 15$ (a drop of $12$ ), and the answer probably "nice", guess it lands near $13$ again.

Same minus same — watch the double negatives. $x$ : $2 - (- 1) = 2 + 1 = 3$ ; $y$ : $- 2 - 2 = - 4$ ; $z$ : $- 15 - (- 3) = - 15 + 3 = - 12$ . Why this step? Subtracting a negative adds. The $y$ and $z$ gaps come out negative — that is fine, because the next step squares them away.
Square, add. $3^{2} + (- 4)^{2} + (- 12)^{2} = 9 + 16 + 144 = 169$ . Why this step? $(- 4)^{2} = + 16$ and $(- 12)^{2} = + 144$ : squaring is order- and sign-blind, so distance never turns negative.
Root. $169 = 13$ .

Answer: $13$ units.

Verify: Reverse the subtraction order: $x_{1} - x_{2} = - 1 - 2 = - 3$ , and $(- 3)^{2} = 9$ — identical. Order does not change a squared term.

Cell C — a point on an axis (zeros)

Worked example Example 3 — a coordinate is zero

Find the distance from $A (0, 0, 4)$ to $B (3, 4, 0)$ . $A$ sits on the $z$ -axis; $B$ sits in the $x y$ -plane.

Forecast: One point is "up the pole", the other "on the floor". Expect the diagonal to be a bit longer than either the height $4$ or the floor-diagonal $5$ .

Same minus same. $x$ : $3 - 0 = 3$ ; $y$ : $4 - 0 = 4$ ; $z$ : $0 - 4 = - 4$ . Why this step? A zero coordinate is not special — it just means that term equals the other point's coordinate. Nothing vanishes here, but a term would vanish if both points shared a coordinate.
Square, add. $3^{2} + 4^{2} + (- 4)^{2} = 9 + 16 + 16 = 41$ .
Root. $41 \approx 6.40$ . Why this step? $41$ is not a perfect square, so leave it as $41$ — exact beats a rounded decimal in an exam.

Answer: $41$ units $\approx 6.40$ .

Verify: $6^{2} = 36$ and $7^{2} = 49$ , so $41$ lies between $6$ and $7$ — matches the forecast that it's a touch above the floor-diagonal $5$ .

See the two special points drawn in the room:

Cell D — degenerate: the same point

Cell E — solve for an unknown coordinate

Worked example Example 5 — find the missing

$y$ (two answers) Point $P (3, y, 4)$ is at distance $5$ from $Q (3, 1, 0)$ . Find $y$ .

Forecast: $x$ matches ( $3 = 3$ ) so that term is dead. The $z$ -gap is $4$ . To reach a total of $5$ , the $y$ -part must supply a leg — do you sense $3$ coming?

Write the equation, square both sides first. $5^{2} = (3 - 3)^{2} + (y - 1)^{2} + (4 - 0)^{2}$ Why this step? Squaring removes the outer root, turning a scary radical equation into ordinary algebra.
Simplify the known terms. $25 = 0 + (y - 1)^{2} + 16$ , so $(y - 1)^{2} = 9$ . Why this step? Isolate the unknown before rooting.
Root — and keep both signs. $y - 1 = \pm 3$ , so $y = 4$ or $y = - 2$ . Why this step? $9$ has two roots $+ 3$ and $- 3$ ; a distance condition is symmetric, so two positions (one above, one below) are equally valid.

Answer: $y = 4$ or $y = - 2$ .

Verify: $y = 4$ : $0 + 9 + 16 = 25 = 5$ ✓. $y = - 2$ : $0 + 9 + 16 = 5$ ✓. Both check out — never discard a root without a reason.

Cell F — geometry proof (triangle type)

Worked example Example 6 — right-angled triangle

Show that $A (1, 2, 3)$ , $B (3, 4, 3)$ , $C (3, 2, 5)$ form a right-angled triangle.

Forecast: If it's right-angled, two short sides squared should add to the longest side squared — Pythagoras in reverse.

Compute all three side lengths (squared — root later). $A B^{2} = (3 - 1)^{2} + (4 - 2)^{2} + (3 - 3)^{2} = 4 + 4 + 0 = 8$ . $B C^{2} = (3 - 3)^{2} + (2 - 4)^{2} + (5 - 3)^{2} = 0 + 4 + 4 = 8$ . $A C^{2} = (3 - 1)^{2} + (2 - 2)^{2} + (5 - 3)^{2} = 4 + 0 + 4 = 8$ . Why this step? Comparing squared lengths avoids messy square roots and is exactly what Pythagoras needs.
Test Pythagoras. Is any pair's sum equal to the third? $A B^{2} + B C^{2} = 8 + 8 = 16$ , but $A C^{2} = 8$ . Not right-angled that way. Every pair gives $8 + 8 = 16 \neq = 8$ . Why this step? All three sides are equal ( $8$ each), so no side is a hypotenuse.
Reinterpret. Since $A B = B C = C A = 8 = 22$ , the triangle is equilateral, not right-angled. Why this step? Honest maths: the data told us "equilateral", so we report the truth. (An exam may deliberately mislabel to test you.)

Answer: The triangle is equilateral with side $22$ , not right-angled.

Verify: All three squared distances equal $8$ ✓, so all sides equal — the defining property of an equilateral triangle.

Cell G — real-world word problem

Worked example Example 7 — ant in a room

A room is a box $6 m$ long, $4 m$ wide, $3 m$ tall. An ant sits at a bottom corner; a crumb sits at the diagonally opposite top corner. What is the straight-line gap?

Forecast: The straight gap must beat the floor-diagonal alone ( $6^{2} + 4^{2} = 52 \approx 7.2$ ). Guess something a little over $7$ .

Set coordinates. Put the ant at the origin $(0, 0, 0)$ ; then the opposite top corner is $(6, 4, 3)$ . Why this step? Choosing the ant as origin makes every coordinate of the crumb equal to a room dimension — the word problem becomes a plug-in.
Apply the formula. $gap = 6^{2} + 4^{2} + 3^{2} = 36 + 16 + 9 = 61$ . Why this step? Length, width, height are exactly the box edges $a, b, c$ ; the body diagonal is the straight-line distance.
Numeric value. $61 \approx 7.81 m$ .

Answer: $61 \approx 7.81$ metres.

Verify: $7^{2} = 49 < 61 < 64 = 8^{2}$ , so the gap is between $7$ and $8$ m, and indeed a bit above the floor-diagonal $7.2$ m ✓.

Cell H — exam twist: equidistant point

Worked example Example 8 — the locus twist

Find the point on the $z$ -axis that is equidistant from $A (1, 2, - 1)$ and $B (3, - 2, 1)$ .

Forecast: A point on the $z$ -axis looks like $(0, 0, z)$ . "Equidistant" means one equation — expect a single $z$ .

Name the unknown point. Any point on the $z$ -axis is $P (0, 0, z)$ (its $x$ and $y$ are zero by definition of the axis). Why this step? Encoding "on the $z$ -axis" as $(0, 0, z)$ turns a wordy condition into one variable.
Set $P A^{2} = P B^{2}$ (square both, no roots). $(0 - 1)^{2} + (0 - 2)^{2} + (z + 1)^{2} = (0 - 3)^{2} + (0 + 2)^{2} + (z - 1)^{2}$ Why this step? "Equidistant" means $P A = P B$ ; squaring both sides removes both roots at once and keeps the equation linear-friendly.
Expand and cancel. Left: $1 + 4 + z^{2} + 2 z + 1 = z^{2} + 2 z + 6$ . Right: $9 + 4 + z^{2} - 2 z + 1 = z^{2} - 2 z + 14$ . Set equal: $z^{2} + 2 z + 6 = z^{2} - 2 z + 14$ . The $z^{2}$ terms cancel, giving $4 z = 8$ , so $z = 2$ . Why this step? The squared- $z$ terms always cancel in an "equidistant" set-up, leaving a simple linear equation — this is the signature trick.

Answer: $P (0, 0, 2)$ .

Verify: $P A^{2} = 1 + 4 + (2 + 1)^{2} = 1 + 4 + 9 = 14$ . $P B^{2} = 9 + 4 + (2 - 1)^{2} = 9 + 4 + 1 = 14$ . Equal ✓ — $P$ really is equidistant.

Recall

Recall Which cases are "green" now?

Positive plug-in ::: Ex 1 Negative / double-negative coords ::: Ex 2 Coordinate equal to zero ::: Ex 3 Coincident points (distance $0$ ) ::: Ex 4 Unknown coordinate → two answers ::: Ex 5 Triangle-type proof via three distances ::: Ex 6 Word problem (box diagonal) ::: Ex 7 Equidistant point → linear equation after cancelling squares ::: Ex 8

Recall Quick numeric checks

Distance $(2, 3, 6) \to (5, 7, 18)$ ::: $13$ Distance $(- 1, 2, - 3) \to (2, - 2, - 15)$ ::: $13$ Distance $(0, 0, 4) \to (3, 4, 0)$ ::: $41$ If $(3, y, 4)$ is $5$ from $(3, 1, 0)$ ::: $y = 4$ or $y = - 2$ Ant-to-crumb in a $6 \times 4 \times 3$ box ::: $61 \approx 7.81$ Point on $z$ -axis equidistant from $(1, 2, - 1)$ and $(3, - 2, 1)$ ::: $(0, 0, 2)$

Connections

Distance formula in 3D — the parent rule every example uses.
Distance formula in 2D — Ex 3's floor-diagonal is exactly this special case.
Vectors and magnitude — each distance is $∣ P Q ∣$ .
Section formula in 3D — shares the coordinate-difference machinery.
Coordinate axes and octants in 3D — Ex 3 and Ex 8 place points on axes.
Equation of a sphere — Ex 8's "equidistant" idea generalises to a plane/sphere locus.

Worked examples — Distance formula in 3D

The scenario matrix

Cell A — plain plug-in

Cell B — negative coordinates

Cell C — a point on an axis (zeros)

Cell D — degenerate: the same point

Cell E — solve for an unknown coordinate

Cell F — geometry proof (triangle type)

Cell G — real-world word problem

Cell H — exam twist: equidistant point

Recall

Connections

Concept Map