01 · A coordinate plane
Which coordinate plane contains (4, 0, −7)?
Hint
Identify the coordinate which is zero.
Worked solution
The xz-plane, because y = 0.
Understand · explore · practise
Coordinates and distance in 3D
Read three-dimensional coordinates and calculate distances using Pythagoras. Find unknown coordinates, handle negative components, locate midpoints and distinguish a space diagonal from a surface route.
Before you startPythagoras in two dimensions, signed coordinates and square roots.
01 / Add a third perpendicular coordinate
A point needs a position along each of three axes.
P = (x, y, z)
The x-, y- and z-axes are mutually perpendicular in the coordinate system.
Think of x and y locating a point on a horizontal plane, with z recording its signed height. A flat sketch is only a projection of this three-dimensional arrangement. The order of the coordinates matters.
Three perpendicular changes, one distanceExplore
P = (3, 4, 2).
Horizontal distance squared = 3² + 4² = 25.
OP² = 25 + 2² = 29; OP = √29 ≈ 5.385165.
Amber shows the three component steps; blue shows the straight displacement. Changing the view changes only the drawing. Screen lengths and angles are projections; calculate the true distance from the coordinates.
02 / Interpret zero and negative coordinates
Zero fixes a point on a coordinate plane.
Locate P = (−2, 5, 0) and Q = (0, 0, −3).Worked example
P lies in the xy-plane
Its z coordinate is zero.
P has negative x and positive y
Move in the indicated signed directions.
Q lies on the negative z-axis
Both horizontal coordinates are zero.
02 · A coordinate axis
Describe the position of (0, −6, 0).
Hint
Only one coordinate is nonzero.
Worked solution
It lies six units along the negative y-axis.
03 / Apply Pythagoras twice
Combine the horizontal distance with the vertical change.
Find the distance from O to P = (3, 4, 12).Worked example
Horizontal distance² = 3² + 4² = 25
First work in the xy-plane.
OP² = 25 + 12² = 169
The vertical segment is perpendicular to that plane.
OP = 13
Distance is the nonnegative square root.
Watch: two right triangles give a space distance
Pause, replay or seek freely. The notes explain the same idea and stay in view.
03 · Another distance from the origin
Find the distance from O to (−2, 3, 6).
Hint
Square all three signed components.
Worked solution
√[4 + 9 + 36] = 7.
04 / Use changes in coordinates between two points
Subtract corresponding coordinates before squaring.
AB = √[(x₂ − x₁)² + (y₂ − y₁)² + (z₂ − z₁)²]
Each bracket is a signed displacement component; their squares give the distance squared.
Find the distance between A = (−1, 2, 5) and B = (3, −1, −7).Worked example
B − A = (4, −3, −12)
Endpoint minus startpoint gives the displacement.
AB² = 4² + (−3)² + (−12)² = 169
Use all three components.
AB = 13
The distance is a scalar, not a three-component vector.
04 · Same height
Find the distance between (1, 2, 4) and (4, 6, 4).
Hint
The vertical difference is zero.
Worked solution
√(3² + 4² + 0²) = 5.
05 · One changing coordinate
Find the distance between (−2, 3, −5) and (−2, 3, 4).
Hint
Only the z coordinate changes.
Worked solution
The distance is |4 − (−5)| = 9.
05 / Keep brackets around negative components
Changing direction changes displacement, but not distance.
Reverse the journey from A to B in the previous example.Worked example
A − B = (−4, 3, 12)
All displacement components reverse sign.
BA² = (−4)² + 3² + 12² = 169
Squaring removes those sign changes.
BA = AB = 13
The two directed displacement vectors are opposites, but the lengths agree.
06 · A sign trap
Why is (−5)² equal to 25 while −5² is normally interpreted as −25?
Hint
Exponentiation happens before the leading unary minus unless brackets group it.
Worked solution
(−5)² multiplies −5 by itself. The expression −5² means −(5²). Use brackets around each negative component.
07 · Signs and length
Do (2, −3, 6) and (−2, 3, −6) have the same distance from the origin?
Hint
Compare the squared components.
Worked solution
Yes. Both distances are 7, though they point in opposite directions.
06 / Turn a distance condition into an equation
A squared difference can give two possible coordinates.
A = (1, −2, 3), B = (4, 2, k), and AB = √41. Find k.Worked example
3² + 4² + (k − 3)² = 41
Square the distance equation.
(k − 3)² = 16
Subtract the fixed contribution 25.
k − 3 = ±4
Retain both square-root signs.
k = 7 or k = −1
Both points are the same distance from A.
08 · A repeated solution
With the same A and B but AB = 5, find k.
Hint
The horizontal changes already contribute 25.
Worked solution
(k − 3)² = 0, so k = 3 only.
09 · No real point
Could the same A and B have AB = 4?
Hint
The horizontal distance is already 5.
Worked solution
No. The equation would require (k − 3)² = −9.
07 / Average each coordinate for a midpoint
The third coordinate follows the same rule as the first two.
Find the midpoint of A = (−2, 4, 1) and B = (6, −2, 7).Worked example
M = ((−2 + 6)/2, (4 − 2)/2, (1 + 7)/2)
Average corresponding coordinates.
M = (2, 1, 4)
This lies halfway along the straight segment.
M − A = (4, −3, 3) = B − M
Equal displacement halves verify the midpoint.
10 · Recover an endpoint
A = (1, 0, −3) and the midpoint M = (4, −2, 1). Find B.
Hint
Use B = 2M − A componentwise.
Worked solution
B = (7, −4, 5).
08 / A space diagonal is a straight path through the solid
Three perpendicular edge lengths determine its length.
A rectangular box has perpendicular side lengths 2, 3 and 6 cm.Worked example
Face diagonal across 2 by 3 = √13 cm
Apply Pythagoras on the base.
Space diagonal = √(13 + 36) = 7 cm
Include the third perpendicular edge.
Following three edges gives 2 + 3 + 6 = 11 cm
That is a different, longer route.
11 · A cube diagonal
Find the space diagonal of a cube with side length 2 cm.
Hint
All three component lengths are 2.
Worked solution
√(4 + 4 + 4) = 2√3 cm.
12 · Infer an edge length
A rectangular box has two perpendicular edge lengths 3 and 4 cm and space diagonal 13 cm. Find the third edge.
Hint
Its square is 13² − 3² − 4².
Worked solution
The third edge is √144 = 12 cm.
09 / A surface route has a different constraint
Unfold adjacent faces before applying two-dimensional Pythagoras.
An ant travels between opposite vertices of an axis-aligned cube of side 2 cm, staying on its surface.Worked example
The straight space diagonal passes through the cube
It is not an allowed surface route.
Unfold two adjacent faces into a 4 by 2 rectangle
A straight line across the unfolding gives a surface route.
Length = √(4² + 2²) = 2√5 cm
The three choices of paired faces are symmetric for a cube; this is the shortest surface route.
13 · Compare routes
For this cube, compare the space diagonal, shortest surface route and a three-edge route.
Hint
The constraints differ.
Worked solution
2√3 < 2√5 < 6 cm. A shortest path depends on where travel is allowed.
10 / Check scale, units and the size of the answer
A distance cannot be smaller than any component change.
Coordinate differences must use the same length unit. Keep square roots exact unless a decimal answer is requested. The distance is at least the absolute value of each individual change; this gives a quick check on arithmetic.
14 · A useful bound
Two points differ by 8 in z. Could their distance be 6?
Hint
The other squared changes are nonnegative.
Worked solution
No. Their distance is at least 8.
15 · Convert before calculating
Three perpendicular displacements are 3 m, 400 cm and 12 m. Find the distance.
Hint
Convert 400 cm to 4 m.
Worked solution
√(3² + 4² + 12²) = 13 m.
11 / Subtract, square, add and take the nonnegative root
Coordinates locate points; differences describe a displacement.
- Read coordinates in x, y, z order.
- Use zero coordinates to identify planes and axes.
- Apply Pythagoras to three perpendicular changes.
- Keep brackets around negative values.
- Retain both roots for an unknown coordinate when appropriate.
- Distinguish straight distance, edge paths and surface routes.
- Use matching units and sensible bounds.
16 · Final distance
Find the exact distance from (2, −1, 4) to (−1, 3, 6).
Hint
The displacement components are −3, 4 and 2.
Worked solution
The distance is √29.
Section 1 of 11 · Add a third perpendicular coordinate