Shrink (or stretch) an arrow to length 1: A unit vector points in a chosen direction but is exactly 1 unit long. It's like a compass needle: it tells you which way, not how far.
How to build one: To build the unit vector pointing the same way as v, divide v by its own length |v|. Dividing by the length scales every component down so the new length comes out to exactly 1.
IB-style question — find a unit vector
Find the unit vector in the direction of a = (3, 4)ᵀ.
Step by step
- First find the length, so you know what to divide by.
- Divide each component by the length 5.
- Check: its length should be 1.
Final answer
â = (3/5, 4/5)ᵀ (or 0.6i + 0.8j).
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From the origin to a point, and from point to point: The position vector of a point A is the arrow from the origin O straight to A, written OA (its components are just A's coordinates).
From A to B
- First travel back to the origin and then out to B: AB = OB − OA ("finish minus start").
- Its length |AB| is the straight-line distance between A and B.
IB-style question — the vector joining two points
Points A and B have position vectors OA = (1, 2)ᵀ and OB = (4, 6)ᵀ.
Find AB and its magnitude.
Step by step
- 'Finish minus start': subtract OA from OB, component by component.
- The magnitude of AB is the distance between A and B.
Final answer
AB = (3, 4)ᵀ, |AB| = 5.
IB-style question — distance in 3D
Find the distance between P(2, −1, 3) and Q(4, 1, −1).
Step by step
- Form PQ = OQ − OP (finish minus start).
- The distance is |PQ|.
- Simplify the surd: √24 = √(4·6) = 2√6.
Final answer
Distance PQ = 2√6 ≈ 4.90.