Distance formula in 3D: Question bank — Maths deep dive

Before we start, one word we will lean on: a degenerate case just means an input pushed to an extreme — two points sitting on top of each other, or all on one line, or one coordinate frozen to zero. These edges are where formulas quietly break, so we test them on purpose.

True or false — justify

(x_{2} - x_{1})

and

(x_{1} - x_{2})

give different distances.

False — each difference is squared, and

(- k)^{2} = k^{2}

, so the order is invisible in the answer. The two subtractions differ only in sign, which the square erases.

The distance

P Q

can be negative if

Q

is "below"

P

False — it is a square root of a sum of squares, and each square is

\geq 0

, so

P Q \geq 0

always. "Below" only flips a sign inside a bracket, which squaring removes.

If two points share the same

x

and

y

but differ in

z

, the formula reduces to

∣ z_{2} - z_{1} ∣

True — the

x

and

y

brackets are zero, leaving

(z_{2} - z_{1})^{2} = ∣ z_{2} - z_{1} ∣

. The square root of a single square is the absolute value, not just

z_{2} - z_{1}

P Q = 0

is impossible for two genuine points.

False —

P Q = 0

exactly when all three brackets vanish, i.e.

P

and

Q

are the same point. A sum of squares is zero only if every term is zero.

Swapping the roles of

P

and

Q

changes the distance.

False — swapping negates every bracket, and squaring undoes that, so

P Q = QP

. Distance is symmetric by construction.

The 2D distance formula is just this 3D one with

z_{1} = z_{2}

True — equal

z

makes the third bracket zero, collapsing to

(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

, the 2D formula. 3D genuinely contains 2D as a flat slice.

For points on the same coordinate axis, distance equals the plain difference of their non-zero coordinates.

True in magnitude — two other brackets are zero, so you get the absolute value of the surviving difference. Still an

∣ \cdot ∣

, so never negative.

a^{2} + b^{2} + c^{2}

can be simplified to

a + b + c

False —

a^{2} + b^{2} + c^{2} \neq = a + b + c

in general (try

a = b = c = 1

3 \approx 1.73

3

). Square-root of a sum is not the sum of roots.

Spot the error

P Q = (x_{2} - x_{1}) + (y_{2} - y_{1}) + (z_{2} - z_{1})

The squares and the outer root are gone. Adding legs is only valid along a single axis; a true straight-line gap is a hypotenuse, so it must be square-add-root.

P Q = (x_{2} - y_{1})^{2} + (y_{2} - z_{1})^{2} + (z_{2} - x_{1})^{2}

Coordinates are cross-matched. You must subtract same-named coordinates:

x

with

x

y

with

y

z

with

z

— mixing them measures nothing geometric.

"Distance from origin of

(2, - 3, 6)

2^{2} - 3^{2} + 6^{2}

A minus sign leaked in from the raw coordinate

- 3

. Every term is a square, hence positive; it should be

2^{2} + (- 3)^{2} + 6^{2} = 49 = 7

P Q^{2} = 169

, so

P Q = 169

The final square root was dropped.

P Q = 169 = 13

; the formula always ends with the outer root undoing the sum of squares.

(x_{2} - x_{1})^{2} = x_{2} - x_{1}

always."

Only true when

x_{2} \geq x_{1}

; in general

(x_{2} - x_{1})^{2} = ∣ x_{2} - x_{1} ∣

. Forgetting the absolute value flips signs in axis-aligned distances.

"We applied Pythagoras once to get the body diagonal."

It takes twice — once for the floor diagonal

d^{2} = a^{2} + b^{2}

, then again with the vertical leg

c

to reach

P Q^{2} = d^{2} + c^{2}

. Skipping a step loses the

c^{2}

term.

"To prove

A, B, C

collinear, just check

A B = B C

Equal short distances only means

B

is a midpoint-ish balance, not that the three lie on a line — they could form an isosceles triangle. Collinearity needs the largest distance to equal the sum of the other two.

Why questions

Why does squaring appear in the formula at all?

Because distance is the hypotenuse of a right triangle, and Pythagoras relates squares of the sides — the squaring is inherited from

a^{2} + b^{2} = c^{2}

, applied twice.

Why is Pythagoras allowed twice — what makes each triangle right-angled?

The

x

y

z

axes are mutually perpendicular, so the floor edges meet at

9 0^{\circ}

, and the vertical

z

-leg is perpendicular to the whole floor (hence to the floor diagonal too). Two clean right angles, two Pythagoras uses.

Why don't we need absolute-value bars

∣ \cdot ∣

in the final formula?

Because each bracket is squared, and squaring already discards sign. The

∣ \cdot ∣

we'd use on individual edges becomes redundant once everything is squared.

Why is

P Q

the same as the magnitude of the vector

P Q

Because

P Q = (x_{2} - x_{1}, y_{2} - y_{1}, z_{2} - z_{1})

and its length (see Vectors and magnitude) is

sum of squared components

— literally the same expression. Distance and vector length are two names for one quantity.

Why does the formula treat all three axes identically?

Because space is isotropic in this coordinate system — no axis is special, so

x

y

z

enter symmetrically. That symmetry is why the sphere built from it is perfectly round.

Why can the same distance equation give two answers when solving for an unknown coordinate?

Because after squaring you meet

(x - a)^{2} = k

, which has two roots

x = a \pm k

. Geometrically, two points on a line can sit equally far on either side of the reference point.

Edge cases

What does the formula give for two identical points

P (3, 3, 3)

and

Q (3, 3, 3)

Zero — every bracket is

0

, so

P Q = 0 = 0

. This is the only way a distance can vanish, and it correctly says "no gap".

If all three points lie exactly on the

z

-axis, is collinearity even a question?

They are automatically collinear — sharing

x = y = 0

forces them onto one line, so the "sum of distances" test will pass trivially. The formula still gives correct pairwise gaps as differences of

z

Two points differ only in a single coordinate — does 3D distance "waste" the other two terms?

Not wasted — the unused brackets are simply

0

and drop out, leaving the plain axis distance. The formula gracefully degenerates to a 1D difference.

What happens to the formula if a point sits at the origin?

Then

x_{1} = y_{1} = z_{1} = 0

, and it collapses to

O P = x^{2} + y^{2} + z^{2}

using only the other point's coordinates. The origin case is not special — it's the general formula with one corner pinned to

(0, 0, 0)

Can three distinct points give

A B + B C = A C

yet not be collinear?

No — equality in the triangle inequality forces

B

onto the straight segment

A C

. Strict collinearity and that equality are equivalent for distinct points.

If a point lies on a coordinate plane (say

z = 0

), does anything in the formula change?

No —

z = 0

just makes that point's

z

-term equal

- z_{1}

of the other, nothing degenerates. Lying on a plane is not an edge case for distance, only for which octant you're in.

Is there any input for which the square root is undefined?

No — the argument

(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} + (z_{2} - z_{1})^{2}

is a sum of squares, always

\geq 0

, so the real square root is always defined. The formula never fails on valid coordinates.

Connections

Distance formula in 3D — the parent this bank stress-tests.
Distance formula in 2D — the $z \to$ flat degenerate case above.
Vectors and magnitude — why $P Q = ∣ P Q ∣$ .
Section formula in 3D — same coordinate differences, used for internal points.
Direction cosines and direction ratios — the differences, normalised.
Equation of a sphere — the "all points at fixed distance" idea.
Coordinate axes and octants in 3D — where the sign cases live.