Maths › Vectors › Vectors in three dimensions
Vectors in three dimensions
Add a third unit vector, k, and vectors leave the page. Almost nothing else changes: magnitudes still come from Pythagoras, now applied twice, AB is still b minus a, parallel still means scalar multiple, and the geometry of boxes, triangles and midpoints in space reduces to the same slot-by-slot arithmetic.
Builds on Vectors in two dimensions.
IN THIS TOPIC
- Work with i, j, k components, magnitudes and distances in three dimensions.
- Solve geometric problems in space: collinearity, midpoints, and triangle shapes by distance alone.
COMMON MISCONCEPTION
Adding a third dimension needs a whole new toolkit.
One more slot
A third unit vector k points out of the old page, and every vector becomes xi + yj + zk, a column of three. The magnitude formula gains a term and stays on the must-learn list.
It is Pythagoras run twice, once across the floor of a box and once up its wall.
WORKED EXAMPLE
Magnitude, unit vector, distance
For a = 2i + 3j + 6k, find |a| and the unit vector in the direction of a; then find the distance between A(1, 2, 3) and B(3, −1, 5).
|a| = √(4 + 9 + 36) = √49 = 7, a pleasingly exact answer from a 2, 3, 6 triple.
The unit vector is a/7 = (2/7)i + (3/7)j + (6/7)k.
= b − a = 2i − 3j + 2k, so the distance is |AB| = √(4 + 9 + 4) = √17.
Every move here was the 2D lesson with one extra slot. The toolkit came along unchanged.
GUIDED PRACTICE
Scaling to order
For p = 6i − 2j + 3k, find |p|, the unit vector in the direction of p, and a vector of magnitude 21 parallel to p, before opening the working.
Show the working
|p| = √(36 + 4 + 9) = 7, another exact triple.
The unit vector is (6/7)i − (2/7)j + (3/7)k.
21 = 3 × 7, so 3p = 18i − 6j + 9k works, and −3p points the other way along the same line. Mention both; it is a free mark.
Geometry with components
Position vectors, AB = b − a, parallel meaning scalar multiple: all of it survives the move into space. Three points are collinear when the vectors joining them are scalar multiples of each other. Take P(1, 2, 3), Q(3, 3, 5) and R(7, 5, 9). Then PQ = (2, 1, 2) and PR = (6, 3, 6) = 3PQ, so the three sit on one line, with Q a third of the way from P to R. Midpoints still average position vectors, slot by slot.
WORKED EXAMPLE
Classifying a triangle by distances
The points A(2, 1, 3), B(4, 2, 5) and C(0, 3, 4) form a triangle. Show it is right angled and isosceles.
AB = (2, 1, 2) with |AB| = 3, and AC = (−2, 2, 1) with |AC| = 3, so the triangle is isosceles.
BC = (−4, 1, −1) with |BC| = √18 = 3√2.
|AB|2 + |AC|2 = 9 + 9 = 18 = |BC|2, so by the converse of Pythagoras the angle at A is right.
Distances alone settled it. In standard A-level Maths there is no dot product to reach for (Further Maths students meet one in Core Pure, and should still hold it back in a 9MA0 paper), so distances and the converse of Pythagoras are the route for every triangle question in this unit.
INDEPENDENT PRACTICE
A parallelogram in space
ABCD is a parallelogram with A(1, 0, 2), B(3, 1, 4) and C(6, 3, 5). Find the coordinates of D.
Show the working
As in two dimensions, = forces d = a + c − b.
d = (1 + 6 − 3, 0 + 3 − 1, 2 + 5 − 4), so D is (4, 2, 3).
Check: DC = (2, 1, 2) = AB, so the shape closes, and the third coordinate rode along without complaint.
ASSESSMENT FOCUS
- Magnitude is the square root of the sum of three squares. The 2, 3, 6 and 1, 2, 2 style triples land exact, and papers love them.
- Collinearity is parallelism through a shared point. Write the scalar down and name the common point.
- Keep components in a column as working. A sign slip in the middle slot is this topic's commonest lost mark.
CHECK YOURSELF
P is (2, −1, 4) and Q is (4, 3, 8). Find |PQ|, the midpoint of PQ, and a unit vector parallel to PQ.
Show a hint
PQ = q − p first; the magnitude is a whole number.
Show the answer
PQ = 2i + 4j + 4k, and |PQ| = √(4 + 16 + 16) = 6, the 1, 2, 2 triple scaled by 2.
The midpoint averages the position vectors: (3, 1, 6).
Dividing by the length, (1/3)i + (2/3)j + (2/3)k, with its negative equally valid the other way along the line.
A third slot, the same rules: magnitudes by three-square Pythagoras, AB = b − a as ever.
Space geometry reduces to distances and scalar multiples, checked component by component.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their worked answers on the vectors in three dimensions questions page.
CHECK YOUR PROGRESS
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- Work with i, j, k components, magnitudes and distances in three dimensions.
- Solve geometric problems in space: collinearity, midpoints, and triangle shapes by distance alone.
Open the full revision checklist to see every objective in the course in one place.