Maths › Vectors › Vectors in two dimensions
Vectors in two dimensions
Some quantities need a direction as well as a size, and vectors carry both in one object. Arrows add tip to tail, components add slot by slot, and position vectors turn every geometry problem about points and midpoints into arithmetic you can do in the margin.
Builds on Straight lines and Triangles and the sine and cosine rules.
Where it earns its keep: Scalars and vectors on InkPhysics.
IN THIS TOPIC
- Work in column and i-j notation, converting between components and magnitude-direction form.
- Add vectors, scale them, and recognise parallel vectors.
- Use position vectors, = b − a, and distances to solve geometric problems.
- Find the unit vector in the direction of a given vector, and scale it to any length asked for.
COMMON MISCONCEPTION
The magnitude of a + b is |a| + |b|.
Arrows with components
A vector carries magnitude and direction together, and two notations hold it. Column form stacks the components. The i-j form writes multiples of the unit vectors i, one step across, and j, one step up. Magnitude |a| is Pythagoras on the components; direction comes from trigonometry. So (3, 4) has magnitude 5 and points 53.1° above the horizontal. Divide any vector by its own magnitude and you are left with a unit vector, length 1, same direction, which is the building block for every “find a vector of magnitude 39” question the papers set.
WORKED EXAMPLE
Between the two forms
Find the magnitude and direction of a = 3i + 4j, and the unit vector in the direction of a.
|a| = √(32 + 42) = 5.
Direction: tan θ = 4/3, so θ = 53.1° above the positive x-direction.
The unit vector is a divided by its length, (3/5)i + (4/5)j, whose own magnitude is 1 by construction.
Going the other way, a vector of magnitude 6 at 30° has components (6 cos 30°, 6 sin 30°) = (3√3, 3). Same vector, different notation, and questions swap between them freely.
Adding, scaling, comparing
Vectors add tip to tail. Walk along a, then along b, and the sum is the single straight walk from start to finish. In components that is addition slot by slot. Multiplying by a scalar stretches a vector without turning it, or reverses it if the scalar is negative, so two vectors are parallel exactly when one is a scalar multiple of the other. 6i − 8j is 2(3i − 4j), and there is your scalar.
The same picture settles the question of magnitudes. |a + b| is the length of the shortcut, and a shortcut is shorter than the detour whenever the two legs point different ways. Take a = (3, 4) and b = (4, −3). Then |a| + |b| = 10 while |a + b| = √50 ≈ 7.07. Magnitudes only add when the two vectors already share a direction.
GUIDED PRACTICE
Parallel or not
Given p = 2i − 5j and q = −6i + 15j, show that p and q are parallel, and state the ratio of their magnitudes, before opening the working.
Show the working
q = −3p, a scalar multiple, so the two are parallel, pointing opposite ways because the scalar is negative.
|q| = 3|p|, so the ratio is 1 : 3.
The method begins and ends with spotting the scalar. Checking one component pair suggests it; checking both confirms it.
Position vectors and geometry
Fix an origin O and every point acquires a position vector, a for the point A. The vector from A to B is then
destination minus start, and its magnitude is the distance between the two points. One identity, and geometry becomes component arithmetic. Midpoints average position vectors. Shapes close when their side vectors match.
WORKED EXAMPLE
From one point to another
The points A(1, 5) and B(4, 1) have position vectors a and b. Find the vector and the distance |AB|.
= b − a = (4 − 1)i + (1 − 5)j = 3i − 4j.
The distance is |AB| = √(9 + 16) = 5. The arrow is the difference in the notation: is the walk from A to B, direction and all, and |AB| without it is only how far that is.
Destination minus start, always. Subtract the other way and you get BA, the same walk reversed, and every sign downstream flips with it.
INDEPENDENT PRACTICE
Completing the parallelogram
ABCD is a parallelogram with A(2, 1), B(5, 3) and C(9, 2). Find the position vector of D.
Show the working
In a parallelogram = , so b − a = c − d.
Rearranging, d = a + c − b = (2 + 9 − 5)i + (1 + 2 − 3)j = 6i + 0j, the point (6, 0).
A check is built in. = c − d = 3i + 2j, which matches = 3i + 2j, so the shape genuinely closes.
Label order matters. ABCD names the corners in sequence around the shape, and a different order names a different parallelogram with a different fourth corner.
ASSESSMENT FOCUS
- Set out vector working in column or i-j form and stay in it. Mixing notations halfway through a solution breeds sign errors.
- Magnitude is Pythagoras on components; direction is trigonometry, quoted against a stated reference direction.
- Parallel means scalar multiple. Write the scalar down explicitly, sign included.
- In shape problems, translate the geometric fact into a vector equation, = for a parallelogram, before touching any numbers.
CHECK YOURSELF
Given a = 5i − 12j, find |a|, the unit vector in the direction of a, and a vector of magnitude 39 parallel to a.
Show a hint
5, 12 and a familiar hypotenuse; then scale.
Show the answer
|a| = √(25 + 144) = 13.
The unit vector is (5/13)i − (12/13)j.
Magnitude 39 is 3 × 13, so 3a = 15i − 36j works, and so does −3a in the opposite direction. Saying so earns the final mark.
Components add slot by slot, scalars stretch, and parallel means scalar multiple.
= b − a, destination minus start, and its magnitude |AB| is the distance.
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 two dimensions questions page.
CHECK YOUR PROGRESS
Rate how confident you feel with each objective for this lesson. Ratings are saved in this browser, on this device, unless you sign in.
- Work in column and i-j notation, converting between components and magnitude-direction form.
- Add vectors, scale them, and recognise parallel vectors.
- Use position vectors, = b − a, and distances to solve geometric problems.
- Find the unit vector in the direction of a given vector, and scale it to any length asked for.
Open the full revision checklist to see every objective in the course in one place.