Understand · explore · practise
Add and subtract vectors, multiply by a scalar and work with column vectors and i/j notation. Connect component calculations to head-to-tail diagrams.
Before you startCoordinates, negative numbers and linear equations
01 / What a vector describes
A displacement of 4 units east is different from a displacement of 4 units north. A vector keeps both pieces of information. Its magnitude is its length; a scalar, such as a mass or a length on its own, has no direction.
The arrow AB starts at A and ends at B. Two displacement vectors are equal when their lengths and directions agree, even if their arrows are drawn in different places. Translating an arrow preserves the vector; rotating it usually changes the vector.
BA = −AB
Reversing an arrow changes its direction but not its magnitude. Lowercase bold letters such as a will also denote vectors. A vector equation matches the full displacement, not just its length.
02 / Column vectors and i/j
Use a rightward x-direction and an upward y-direction. A vector with horizontal change p and vertical change q can be written as a column, or using the unit vectors i and j:
= pi + qj
i = j =
For example, −2i + 5j means 2 units left and 5 units up. These are changes in position, not automatically the coordinates of an endpoint. If the arrow starts at (3,−1), it ends at (1,4).
A unit vector has magnitude 1. The two vectors i and j are perpendicular, so they give independent horizontal and vertical directions.
03 / Add head to tail
Place the tail of the second arrow at the head of the first, without changing its length or direction. The direct arrow from the starting point to the finishing point is their sum.
AB + BC = AC
+ =
In the model, a = 3i + j stays fixed. Change b’s two components. The faint arrow shows b at the origin; its translated copy completes the head-to-tail route. Adding in the opposite order gives the same resultant: a + b = b + a.
a = (3,1), b = (1,3). a + b = (4,4). Blue is a; green is b placed at its head; gold is the resultant. The faint dashed arrow is the original b at the origin.
Pause, replay or seek freely. The notes explain the same idea and stay in view.
04 / Subtract a vector
a − b = a + (−b)
For a = 3i + j and b = i + 3j:
a − b = =
Switch the model to a − b: the second head-to-tail arrow is −b, while the faint reference arrow remains b. If a and b are drawn from the same origin to A and B, then the arrow from B to A is a − b. Reversing that order gives b − a.
Subtract the whole component, including its sign. For example, 5 − (−2) = 7, not 3.
05 / Scale and reverse
Multiplying a vector by a scalar λ multiplies each component by λ. For a = 4i − 2j:
½a =
−2a =
A positive multiplier keeps the direction. A negative multiplier reverses it. The magnitude is multiplied by |λ|, so multiplying by −2 doubles the length.
|λa| = |λ| |a|
At λ = 0, the result is the zero vector: no displacement and no defined direction. Also a + (−a) = 0. Do not try to assign a bearing to the zero vector.
06 / Combine operations
Let a = 2i − 3j and b = −i + 4j. To find 3a − 2b:
3a = 6i − 9j
2b = −2i + 8j
Multiply both components in each vector.
3a − 2b = (6 − (−2))i + (−9 − 8)j
= 8i − 17j
The subtraction applies to the whole second vector.
The laws of addition and scalar multiplication behave as expected: λ(a + b) = λa + λb and (λ + μ)a = λa + μa. Multiplying two vectors together is not a component-by-component arithmetic rule in this lesson.
07 / Solve component equations
Suppose a = (p,2) and b = (3,q), written here as component pairs for compactness, and 2a − b = (5,−2). Equate the horizontal and vertical components separately:
2p − 3 = 5 ⇒ p = 4
4 − q = −2 ⇒ q = 6
Let a = 2i + 3j and b = pi − 2j. If a + b is parallel to 4i + j, write a + b = λ(4i + j). The j-component gives λ = 1; the i-component gives 2 + p = 4, so p = 2. The resultant is 4i + j and is non-zero.
To be parallel to i, a non-zero vector needs vertical component 0. To be parallel to j, it needs horizontal component 0. Check that the other component is not also zero before claiming a direction.
08 / Your turn
Write each answer in column form or in i/j notation.
An arrow goes from (−3,2) to (1,−4). Write its vector. Would translating both endpoints 5 units right change it?
Compare the horizontal and vertical changes.
= 4i − 6j
No. Both endpoints move by the same amount, so their difference stays the same.
+
Add matching components.
−
The first horizontal calculation is 4 − (−3).
For a = −2i + 6j, find −½a and describe its length and direction relative to a.
The scalar is negative and has absolute value 1/2.
−½a = i − 3j
Half the length, in the opposite direction.
A route has displacements 5i + j, −2i + 4j and ci + dj. Find c,d if it returns to its start.
The total displacement is the zero vector.
5 − 2 + c = 0 ⇒ c = −3
1 + 4 + d = 0 ⇒ d = −5
a = pi − j, b = 2i + qj and a + 3b = 10i + 8j. Find p,q.
Compare the two components.
p + 6 = 10 ⇒ p = 4
−1 + 3q = 8 ⇒ q = 3
For a = i − 2j and b = −3i + j, find 2a − 3b.
Expand the scalar multiples before subtracting.
2a − 3b = (2 − (−9))i + (−4 − 3)j
= 11i − 7j
For a = i + 4j and b = 2i − j, choose λ so a + λb is parallel to i. State the resultant.
Set the vertical component to zero.
4 − λ = 0 ⇒ λ = 4
a + 4b = 9i
The resultant is non-zero and points right.
A student says a − b and b − a are the same vector because their lengths agree. Explain.
Compare b − a with −(a − b).
They point in opposite directions when non-zero. Equal magnitudes do not imply equal vectors. They are equal only when a = b, making both zero.
09 / Recap
Section 1 of 9 · What a vector describes