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Vectors in two dimensions questions
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.
7 original questions · 22 marks · the vectors in two dimensions notes · Vectors
Every question here is written for this library rather than taken from a past paper. Write your answer out before opening the worked one: the answers award marks point by point, and the marks are easier to see when you have something of your own to compare against.
Find the magnitude of a = 5i + 12j, and the unit vector in the direction of a.
Worked answer
|a| = √(52 + 122) = √169 = 13. The unit vector is a divided by its own length, (5/13)i + (12/13)j. B1 for 13, B1 for the unit vector. Its magnitude is 1 by construction, which makes a quick self-check. Dividing by 13 must reach both components, and halving that job is the usual slip.Given p = 2i − 3j and q = −i + 4j, find 2p + 3q.
Worked answer
2p + 3q = (4 − 3)i + (−6 + 12)j = i + 6j. M1 for multiplying out both brackets, A1 for i + 6j. The arithmetic runs slot by slot, and the only traps are the signs. Multiply through each bracket before collecting.Find the magnitude of 7i − 24j, and the angle it makes with the positive x-direction, to 1 decimal place.
Worked answer
|7i − 24j| = √(49 + 576) = √625 = 25. The angle satisfies tan θ = 24/7, so θ = 73.7°, measured below the positive x-direction because the j-component is negative. B1 for 25, M1 for tan θ = 24/7, A1 for the angle below the axis. Stating which side of the axis the vector lies is part of the answer, since a magnitude and a bare angle do not pin a vector down.The points A(−2, 3) and B(4, −5) have position vectors a and b respectively. Find the vector AB, the distance AB, and the midpoint of AB.
Worked answer
AB = b − a = (4 − (−2))i + (−5 − 3)j = 6i − 8j, destination minus start. The distance is |AB| = √(36 + 64) = √100 = 10. The midpoint averages the position vectors, giving ((−2 + 4)/2, (3 − 5)/2) = (1, −1). M1 for b − a, A1 for 6i − 8j, A1 for the distance, B1 for the midpoint. Writing a − b instead of b − a reverses the vector and costs the first mark, though the distance survives.The vectors 4i + 6j and 6i + λj are parallel. Find the value of λ.
Worked answer
Parallel means one vector is a scalar multiple of the other, so 6i + λj = k(4i + 6j). Comparing i-components gives k = 6/4 = 3/2, and then λ = 6 × 3/2 = 9. Equivalently the components are in proportion, 4/6 = 6/λ, which cross-multiplies to 4λ = 36. M1 for setting up the scalar multiple, A1 for k = 3/2, A1 for λ = 9. Setting up the proportion upside down gives λ = 4, so check the answer against the pattern that both components grow by the same factor.Find the two vectors of magnitude 26 that are parallel to 5i − 12j.
Worked answer
|5i − 12j| = √(25 + 144) = 13, and 26 = 2 × 13, so scale by ±2. The vectors are 10i − 24j and −10i + 24j. M1 for the magnitude 13 and the scale factor 2, A1 for 10i − 24j, A1 for −10i + 24j. Forgetting the negative one loses a mark, because parallel includes pointing the opposite way. A question asking for the same direction rather than parallel would want only the first.Relative to an origin O, the points A and B have position vectors a = i + 6j and b = 16i + 12j. The point P lies on the line segment AB and divides it so that AP : PB = 1 : 2. Find the position vector of P, and hence find |OP|.
Worked answer
P is one third of the way from A to B, so OP = a + ⅓(b − a). Here b − a = 15i + 6j, so ⅓(b − a) = 5i + 2j and OP = (1 + 5)i + (6 + 2)j = 6i + 8j. Then |OP| = √(36 + 64) = 10. M1 for OP = a + ⅓(b − a), A1 for ⅓(b − a), A1 for 6i + 8j, M1 for the magnitude, A1 for 10. Check the ratio afterwards. AP = 5i + 2j and PB = 10i + 4j, which is 2AP, so AP : PB = 1 : 2 as required. The ratio 1 : 2 splits AB into three equal parts, not two, and reading it as a half is the error that dominates this question. The other trap is applying the fraction to b rather than to b − a, which gives a point that does not lie on AB at all. The reliable form is OP = a + [m/(m + n)](b − a) for a ratio m : n.
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