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 and magnitudes via the three-term Pythagoras.
- Find distances, midpoints and unit vectors in three dimensions.
- Solve geometric problems: parallel and collinear checks, and triangle shapes by distances.
WHAT YOU PROBABLY THINK
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,
and it is Pythagoras run twice: once across the floor of a box, 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.
AB = b − a = 2i − 3j + 2k, so the distance is |AB| = √(4 + 9 + 4) = √17.
Every move was the 2D lesson with one extra slot, which is the opening lie dealt with: the toolkit came along unchanged.
YOUR TURN
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, as does −3p pointing the other way, and mentioning both is the safe habit.
Geometry with components
Position vectors, AB = b − a, parallel-means-scalar-multiple: all of it survives the move to space. Three points are collinear when the vectors joining them are scalar multiples: for P(1, 2, 3), Q(3, 3, 5), R(7, 5, 9), PQ = (2, 1, 2) and PR = (6, 3, 6) = 3PQ, so all 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: 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 carried the whole classification; that is the standard route for triangle questions in this course.
TRY IT UNSEEN
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, AB = DC 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.
THE EXAM BIT
- 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.
- AB = b − a in any number of dimensions, destination minus start, and distances are magnitudes of differences.
- Parallel means scalar multiple, and collinearity is parallelism through a shared point; write the scalar down.
- Triangle classification runs on distances and the converse of Pythagoras; no angle formulas are needed or available here.
- Keep components in a column as working; sign slips in the middle slot are 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.
CHECK YOUR PROGRESS
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- Work with i, j, k components and magnitudes via the three-term Pythagoras.
- Find distances, midpoints and unit vectors in three dimensions.
- Solve geometric problems: parallel and collinear checks, and triangle shapes by distances.
No animated video for this topic yet; these notes stand alone.