Position vectors and unit vectors: Additional Maths Practice
Vectors in Two Dimensions
Position vectors and unit vectors
4 self-marking Additional Maths practice questions on Position vectors and unit vectors, written specifically for Cambridge IGCSE Additional Maths (0606) across Single tier. Instant marking with full worked solutions on every question.
Sample Position vectors and unit vectors questions
A few of the 4 questions, from easier to harder. Choose your exam board and tier to practise them, with every answer marked.
Question 12 marksGrade 7
O is the origin. OA=(3−1)
What are the common mistakes in Position vectors and unit vectors?
When a pupil makes one of these on Revision Workshop, the marking engine names the mistake and explains it, instead of just marking the answer wrong.
Adds a fixed amount to each component/quantity to achieve a target change, instead of scaling all components by the same multiplicative factor.
Increasing a vector's magnitude means SCALING it (multiplying every component by the same factor), not adding a fixed amount to each component - since the magnitude needs to triple (from 5 to 15), every component of v is multiplied by 3.
Adds two position vectors together instead of subtracting the starting position vector from the ending one to find a displacement vector.
Quick answers about Position vectors and unit vectors
How many Position vectors and unit vectors practice questions are there?
There are 4 self-marking Additional Maths Maths questions on Position vectors and unit vectors, written for Cambridge IGCSE Additional Maths (0606) across Single tier. Each one is marked instantly and has a full worked solution.
Do I need an account to practise Position vectors and unit vectors?
No. You can start practising Position vectors and unit vectors straight away without an account. A free account keeps your results across sessions, so you can see your progress on this topic over time.
and
OB=(75)
.
Find
AB
. [2]
Show the worked solution
AB=OB−OA=(75)−(3−1)=(46).
Question 23 marksGrade 8
Find the unit vector in the direction of a=(6−8). [3]
Show the worked solution
∣a∣=62+(−8)2=36+64=100=10.
The unit vector is ∣a∣a=101(6−8)=(0.6−0.8)
Question 33 marksGrade 8
The vector v=(34) has magnitude 5.
Find the vector with magnitude 15 that is in the same direction as v. [3]
AB is found by subtracting the STARTING position vector from the ENDING one, OB−OA, not by adding them - (7−35−(−1))=(46).
Fails to divide a vector by its magnitude when finding a unit vector, leaving it unnormalised
A unit vector must have magnitude 1 - dividing each component of a by its magnitude ∣a∣=10 scales it down correctly, giving (0.6−0.8), not the original vector unchanged.