Revision for vector notation, magnitude, unit and position vectors, vector geometry, collinearity, ratios and (0606) velocity problems. Both boards. Cambridge 0606 includes composing velocities and position at time t; Edexcel 4PM1 includes dividing a line in a ratio and simple proofs with vectors.
|ai + bj| = √(a² + b²). The unit vector in the direction of a is a/|a|.
AB = OB − OA ('end minus start').
Parallel vectors are multiples of each other; points A, B, C are collinear if AB and AC are parallel and share A.
If pa + qb = ra + sb and a, b are not parallel, then p = r and q = s.
Position at time t = starting position + t × velocity. Speed is the magnitude of the velocity.
Watch out: Give a magnitude as a positive number, and use the notation the question uses (bold, underlined or with an arrow).
Worked example· 2 marks
The vector \(\mathbf{p} = 8\mathbf{i} - 15\mathbf{j}\). Work out \(|\mathbf{p}|\), and write down the unit vector that points in the same direction as \(\mathbf{p}\).
B1 |p| = 17
B1 (8i − 15j)/17
|p| = √(8² + 15²) = √289 = 17.
The unit vector is p ÷ |p| = (8i − 15j)/17.
Answer: \(|\mathbf{p}| = 17\), unit vector \(\frac{8}{17}\mathbf{i} - \frac{15}{17}\mathbf{j}\)
Vectors questions
Original exam-style questions, written for this site and each re-solved independently. Tags show the courses whose content a question tests and whether it can be done without a calculator (Cambridge 0606 Paper 1 is non-calculator).
Question 1 · 3 marks06064PM1no calculator
Relative to an origin \(O\), \(\overrightarrow{OA} = 2\mathbf{i} + 5\mathbf{j}\) and \(\overrightarrow{OB} = 6\mathbf{i} - 3\mathbf{j}\).
Find \(\overrightarrow{AB}\) and \(|\overrightarrow{AB}|\), giving the magnitude in simplified surd form.
Relative to \(O\), the points \(P\) and \(Q\) have position vectors \(\mathbf{p}\) and \(\mathbf{q}\). The point \(R\) divides \(PQ\) internally in the ratio \(3 : 2\).
Write \(\overrightarrow{OR}\) in terms of \(\mathbf{p}\) and \(\mathbf{q}\).
Show the mark scheme and worked solution
M1 OR = p + (3/5)(q − p)
A1 (2p + 3q)/5
PQ = q − p, and PR is 3/5 of PQ.
OR = OP + PR = p + (3/5)(q − p) = (2/5)p + (3/5)q.
A boat moves with velocity \((6\mathbf{i} + 2\mathbf{j})\) km h⁻¹ relative to the water. The water has velocity \((-2\mathbf{i} + \mathbf{j})\) km h⁻¹. The unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) point east and north.
Find the resultant velocity of the boat, its speed, and the bearing on which it travels.
Show the mark scheme and worked solution
M1 add the velocities
A1 4i + 3j
A1 5 km h⁻¹
A1 bearing 053.1°
Resultant = (6i + 2j) + (−2i + j) = 4i + 3j.
Speed = √(16 + 9) = 5 km h⁻¹.
The bearing is measured clockwise from north: tan θ = 4/3, so θ = 53.13…°, a bearing of 053.1°.
Answer: \(4\mathbf{i} + 3\mathbf{j}\) km h⁻¹, speed \(5\) km h⁻¹, bearing \(053.1^\circ\)
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