IGCSE Additional Mathematics: Vectors in Two Dimensions Practice Questions
A vector in two dimensions is written in component or column form. Its magnitude comes from Pythagoras, a unit vector is the vector divided by its magnitude, and relative motion problems are solved by adding vectors.
Topic 14 of Cambridge IGCSE Additional Mathematics 0606 extends the 0580 treatment into velocity and position vectors. Unit vectors and collinearity proofs are the two techniques that recur. The questions below cover both alongside a relative velocity problem.
What you need to know for Vectors in Two Dimensions
- MagnitudeThe size of a vector, found by Pythagoras on its components. Always positive.
- Unit vectorA vector of magnitude 1 in the same direction, obtained by dividing the vector by its own magnitude.
- ResultantAdd vectors component by component. For perpendicular components, the magnitude follows from Pythagoras and the direction from the inverse tangent.
- Position vectorA vector from the origin to a point. A moving object has position r equal to its initial position plus t multiplied by its velocity.
- Collinear pointsThree points are collinear if one connecting vector is a scalar multiple of another and they share a common point.
IGCSE Additional Mathematics Vectors in Two Dimensions questions and answers
4 exam-style questions written to the 0606 syllabus. Try each one on paper first, then open the worked answer to check your method against the marks.
Find the magnitude of the vector 5i minus 12j, and write down the unit vector in the same direction.
Show the worked answer
- Magnitude = the square root of (52 + (minus 12)2).
- = the square root of (25 + 144) = the square root of 169 = 13.
- A unit vector has magnitude 1 in the same direction, so divide the vector by its magnitude.
- Unit vector = (5i minus 12j) divided by 13, which is 5 over 13 i minus 12 over 13 j.
A boat sets a course due north at 8 km/h. A current flows due east at 6 km/h. Calculate the resultant speed of the boat and the bearing on which it actually travels.
Show the worked answer
- The two velocities are perpendicular, so add them as components: 6 east and 8 north.
- Resultant speed = the square root of (62 + 82) = the square root of 100 = 10 km/h.
- The bearing is measured clockwise from north. The angle east of north satisfies tan theta = 6 divided by 8.
- theta = the inverse tangent of 0.75 = 36.87 degrees.
- The bearing is 036.9 degrees, written with three figures.
A particle starts at the point with position vector 2i + 3j and moves with constant velocity 4i minus j. Find its position vector after 3 seconds.
Show the worked answer
- Position after time t is the initial position plus t multiplied by the velocity.
- r = (2i + 3j) + 3(4i minus j).
- Expand: 3(4i minus j) = 12i minus 3j.
- r = (2 + 12)i + (3 minus 3)j = 14i + 0j, which is 14i. The particle lies on the x-axis at that moment.
The points A, B and C have position vectors i + 2j, 3i + 5j and 7i + 11j. Show that A, B and C are collinear.
Show the worked answer
- Find AB by subtracting the position vector of A from that of B: (3i + 5j) minus (i + 2j) = 2i + 3j.
- Find BC similarly: (7i + 11j) minus (3i + 5j) = 4i + 6j.
- Compare: 4i + 6j = 2(2i + 3j), so BC is a scalar multiple of AB, meaning the two vectors are parallel.
- Since AB and BC are parallel and share the common point B, the three points lie on the same straight line and are therefore collinear.
Common mistakes in this topic
- Giving a negative magnitude.
- Measuring a bearing from the wrong axis.
- Adding a velocity once instead of multiplying by time.
- Concluding collinearity without mentioning the common point.
- Subtracting position vectors in the wrong order when finding a connecting vector.
Exam tips
- Magnitude is always positive. If a negative appears, the squaring step was skipped.
- For a unit vector, divide by the magnitude and leave the answer as a fraction.
- Bearings are measured clockwise from north, with three figures.
- The vector from A to B is the position vector of B minus that of A. Second minus first.
- Finish every collinearity proof with a sentence naming the shared point.
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Vectors in Two Dimensions FAQs
How do I find a unit vector?
Divide the vector by its own magnitude. For 5i minus 12j the magnitude is 13, so the unit vector is 5 over 13 i minus 12 over 13 j. The result always has magnitude 1 and points in the same direction as the original.
How do I find a resultant velocity?
Add the vectors component by component. If the components are perpendicular, the magnitude of the resultant follows from Pythagoras and the direction from the inverse tangent of one component divided by the other, measured from the correct reference direction.
How do I prove three points are collinear?
Find two connecting vectors that share a common point, such as AB and BC. Show that one is a scalar multiple of the other, which makes them parallel. Then state that because they are parallel and share a point, the three points lie on one straight line.
How do I find a position vector after a given time?
Add the initial position vector to the velocity vector multiplied by the elapsed time. The velocity contributes a displacement of velocity times time, so it must be scaled by t rather than simply added once.
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Written to the published Cambridge IGCSE Additional Mathematics (0606) syllabus. Check your school entry code and syllabus year, because Core and Extended candidates are assessed on different content. Last reviewed 2026-08-12.