01 · Read the components
Write 2j − 3k as a three-component vector.
Hint
The i component is zero.
Worked solution
(0, 2, −3).
Understand · explore · practise
Vector components and displacement
Use three-component vectors, distinguish position from displacement, combine journeys head to tail and solve vector equations component by component.
Before you startSigned three-dimensional coordinates and basic vector notation.
01 / Separate position from movement
A point tells you where; a displacement tells you how far in each direction.
AB = b − a
Here a = OA and b = OB are position vectors; bold letters or an arrow distinguish a vector from its length.
In this lesson, a tuple such as (2, −1, 1) used as a vector records the x, y and z components. The same numbers can describe a point relative to the chosen origin. Context and notation tell you which meaning is intended.
Start + displacement = endpointExplore
u = (2, −1, 1); v = (1, 2, 3).
A = (0, 0, 0); B = (2, −1, 1); C = (3, 1, 4).
AB = u; BC = kv = (1, 2, 3); AC = u + kv = (3, 1, 4).
Amber: u. Green: kv. Dashed blue: u + kv. Changing A translates all three points without changing any displacement. At k = 0, B and C coincide. This is a projection: use components, not screen angles or lengths.
02 / Translate between vector notations
The unit vectors i, j and k point along the positive coordinate axes.
v = 3i − 2j + 5k = (3, −2, 5)
As a column vector, write 3, −2 and 5 vertically in that order, inside brackets.
Write −4i + 7k using three components.Worked example
The j component is zero
An omitted term is not a missing coordinate.
v = (−4, 0, 7)
Keep the x, y, z order.
02 · Use unit-vector notation
Write (−1, 4, 0) using i, j, k.
Hint
Multiply each component by the corresponding unit vector.
Worked solution
−i + 4j, or −i + 4j + 0k.
03 / Subtract the start from the end
The sign of each component records the direction of that change.
A = (−2, 3, 1) and B = (4, −1, 6). Find AB.Worked example
AB = (4 − (−2), −1 − 3, 6 − 1)
Subtract corresponding coordinates.
AB = (6, −4, 5)
Move +6 in x, −4 in y and +5 in z.
a + AB = (4, −1, 6) = b
Adding the displacement to the start verifies the endpoint.
03 · Displacement
Find AB for A = (1, −3, 2) and B = (−2, 4, 0).
Hint
Calculate B − A.
Worked solution
AB = (−3, 7, −2).
04 / Reverse all components when reversing a journey
BA = −AB.
Reverse AB = (6, −4, 5).Worked example
BA = (−6, 4, −5)
Every component changes sign.
AB + BA = (0, 0, 0)
The outward and return displacements cancel.
|AB| = |BA| = √77
Direction changes, while length stays the same.
04 · Opposite vector
If PQ = (−3, 7, −2), find QP.
Hint
Negate all three components.
Worked solution
QP = (3, −7, 2).
05 · Zero displacement
What does AB = (0, 0, 0) tell you about A and B?
Hint
No coordinate changes.
Worked solution
A and B are the same point. A zero displacement has no direction.
05 / Recover an unknown endpoint or starting point
An endpoint is a position plus a displacement.
A = (−1, 2, 4) and AB = (3, −5, 2). Find B.Worked example
b = a + AB
Move from the known starting point.
B = (2, −3, 6)
Add componentwise.
06 · Find the start
B = (4, 1, −2) and AB = (−2, 3, 5). Find A.
Hint
a = b − AB.
Worked solution
A = (6, −2, −7).
07 · Position is not displacement
A = (2, 0, 1), B = (3, 4, 1). Why is AB not equal to the position vector b?
Hint
The starting point is not the origin.
Worked solution
AB = b − a = (1, 4, 0), whereas b = (3, 4, 1).
06 / Join journeys head to tail
AB + BC = AC because the intermediate position cancels.
(b − a) + (c − b) = c − a
The −b and +b terms cancel.
u = (2, −1, 1), v = (1, 2, 3). Find u + v.Worked example
u + v = (2 + 1, −1 + 2, 1 + 3)
Add like components.
u + v = (3, 1, 4)
Place v at the head of u to construct the resultant.
Watch: the intermediate point cancels
Pause, replay or seek freely. The notes explain the same idea and stay in view.
08 · Two movements
AB = (1, 3, −2) and BC = (−4, 2, 5). Find AC.
Hint
Add the successive displacements.
Worked solution
AC = (−3, 5, 3).
07 / Subtraction means adding the opposite vector
Be precise about which arrows start at the same point.
OA = a and OB = b. Interpret b − a.Worked example
b − a = AO + OB
AO is the reverse of OA.
AO + OB = AB
The route A → O → B has displacement AB.
a − b = BA
Swapping the order reverses the result.
09 · Subtract components
u = (2, −1, 4), v = (−3, 5, 1). Find u − v.
Hint
Subtract each signed component with brackets.
Worked solution
u − v = (5, −6, 3).
10 · Missing leg
AC = (5, −2, 7), AB = (1, 3, −1). Find BC.
Hint
BC = AC − AB.
Worked solution
BC = (4, −5, 8).
08 / A scalar multiplies every component
Positive scales preserve direction; negative scales reverse it.
λ(x, y, z) = (λx, λy, λz)
For a nonzero vector, length is multiplied by |λ|. At λ = 0 the result is the zero vector.
Scale v = (1, −2, 3) by −2.Worked example
−2v = (−2, 4, −6)
Multiply all three components.
|−2v| = 2√14
A length cannot be negative.
The new vector points oppositely
The negative sign reverses direction.
11 · A linear combination
u = (2, 1, −3), v = (−1, 4, 2). Find 3u − 2v.
Hint
Form each multiple before subtracting.
Worked solution
(6, 3, −9) − (−2, 8, 4) = (8, −5, −13).
12 · Fractional scale
Find −½(6, −4, 2).
Hint
Apply the scale to every component.
Worked solution
(−3, 2, −1).
09 / Equal vectors have equal corresponding components
All three component equations must agree.
Find p and q if p(1, 2, −1) + q(2, −1, 3) = (4, 3, 1).Worked example
p + 2q = 4; 2p − q = 3
Use the first two coordinates.
p = 2, q = 1
Solving the pair gives a candidate.
−p + 3q = 1
The third coordinate agrees, so the candidate solves the vector equation.
13 · Inconsistent third coordinate
Could the same left-hand side equal (4, 3, 2)?
Hint
The first two equations still force p = 2, q = 1.
Worked solution
No. The resulting third component is 1, not 2.
14 · Find an unknown vector
2u + (1, −3, 5) = (7, 1, −1). Find u.
Hint
Subtract the known vector, then divide by 2.
Worked solution
2u = (6, 4, −6), so u = (3, 2, −3).
10 / Changing the origin leaves displacement unchanged
Translate every position vector by the same amount.
Change the origin from O to O′, where OO′ = t.Worked example
a′ = a − t; b′ = b − t
Both position vectors use the new starting origin.
b′ − a′ = (b − t) − (a − t)
The common translation cancels.
b′ − a′ = b − a
AB is independent of the chosen origin.
15 · Shift the origin
A = (1, 2, 3), B = (4, −1, 5), and the new origin has old coordinates (2, 0, −1). Find the new coordinates and AB.
Hint
Subtract (2, 0, −1) from both points.
Worked solution
A′ = (−1, 2, 4), B′ = (2, −1, 6); AB = (3, −3, 2) in either system.
11 / Keep the point, direction and length distinct
Use components to verify every geometric statement.
- Keep x, y, z order, including zero components.
- Displacement is endpoint minus startpoint.
- Reverse a journey by negating every component.
- Add successive journeys head to tail.
- Multiply all components by a scalar.
- Check all three equations when matching vectors.
- A shared translation changes positions but not displacements.
16 · Complete the journey
A = (1, 0, −2), AB = (2, −3, 4), BC = (−1, 5, −2). Find B, C and AC.
Hint
Add each displacement to its startpoint.
Worked solution
B = (3, −3, 2); C = (2, 2, 0); AC = (1, 2, 2).
Section 1 of 11 · Separate position from movement