Pythagoras with a third coordinate: The distance between two points in 3D extends the 2D rule by adding the z-difference: square each coordinate gap, add, square-root.
IB-style question — distance between two points
Find the distance between A(1, 2, 2) and B(4, 6, 14).
Step by step
- Start from the 3D distance formula.
- Substitute A(1, 2, 2) and B(4, 6, 14).
- Work out each gap, then square, add and root.
Final answer
AB = 13.
The exact points from the question: the gaps Δx = 3, Δy = 4, Δz = 12 are the edges of the box, and the space-diagonal AB = √(3² + 4² + 12²) = 13 (not to scale).
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IB-style question — a radius from coordinates
A solid hemisphere has centre A(2, 3, −1), and the point B(6, 0, 11) lies on its curved surface.
Find AB, the radius of the hemisphere.
Step by step
- The radius is the distance from the centre to a point on the surface — start from the 3D distance formula.
- Substitute A(2, 3, −1) and B(6, 0, 11). Watch the double negative: 11 − (−1) = 12.
- Square, add, root.
Final answer
The radius AB = 13.
The solid hemisphere from the question — its radius AB = 13 runs from the centre A to the surface point B (the 3D distance between them).
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Order doesn't matter: Each gap is squared, so (x₂ − x₁) or (x₁ − x₂) gives the same result — the negative disappears.
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Average each coordinate: The midpoint of a segment is found by averaging the x's, the y's and the z's — it sits exactly halfway between the two endpoints.
IB-style question — midpoint
Find the midpoint of A(2, −1, 4) and B(6, 3, 10).
Step by step
- Start from the midpoint formula.
- Substitute A(2, −1, 4) and B(6, 3, 10).
- Simplify each average.
Final answer
Midpoint (4, 1, 7).
IB-style question — centre and radius of a circle
A circle has diameter [AB] with A(−2, 1, 2) and B(6, 1, 8).
Find (a) the centre of the circle and (b) its radius.
Step by step
- (a) The centre is the midpoint of the diameter — start from the midpoint formula.
- Substitute A(−2, 1, 2) and B(6, 1, 8).
- (b) The radius is half the diameter, so first find AB with the distance formula.
- Substitute, then halve.
Final answer
(a) Centre (2, 1, 5). (b) Radius 5.
The circle on diameter [AB] from the question: the centre is the midpoint M(2, 1, 5), and the radius is half of AB = 10, i.e. 5 (not to scale).
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Add then halve: Add the two coordinates and divide by 2 — don't subtract (that would give a gap, not a middle).
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Given a midpoint or distance, find the missing point: If you know the midpoint and one endpoint, set the average equal to the midpoint and solve for the missing coordinate (a missing endpoint is 2M − A).
IB-style question — find the endpoint
M(4, 0, 5) is the midpoint of A(2, −2, 1) and B. Find B.
Step by step
- Rearrange the midpoint relationship: each coordinate of B is 2M − A.
- Substitute M(4, 0, 5) and A(2, −2, 1).
- Simplify.
Final answer
B = (6, 2, 9).
M(4, 0, 5) sits exactly halfway between A(2, −2, 1) and the missing end B. Averaging A and B has to give M back, so each coordinate of B is 2M − A → B(6, 2, 9).
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Check it: Average A and your B — you should get M back.
The space diagonal: The longest distance in a box (the space diagonal) joins two opposite corners — set their coordinates and use the 3D distance formula (or √(l² + w² + h²) for a box).
IB-style question — diagonal of a box
A box has edges 2, 3 and 6 along the axes, one corner at the origin. Find the length of its space diagonal.
Step by step
- The space diagonal joins one corner to the opposite — start from the 3D distance formula.
- One corner is the origin (0, 0, 0) and the opposite corner is (2, 3, 6).
- Evaluate.
Final answer
The space diagonal is 7.
The box from the question: edges 2, 3 and 6 along the axes, one corner at the origin. The space diagonal (corner to opposite corner) is √(2² + 3² + 6²) = 7 (not to scale).
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Same formula, real shape: Whenever you can read off the two corner coordinates, the 3D distance formula does the work.