Questions on vectors in two dimensions

Question 1

Find the magnitude of a = 5i + 12j, and the unit vector in the direction of a.

Accepted 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.

Question 2

Given p = 2i − 3j and q = −i + 4j, find 2p + 3q.

Accepted Answer

2p + 3q = (4 − 3)i + (−6 + 12)j = i + 6j. M1 for multiplying out both brackets, A1 for i + 6j. Multiply through each bracket before collecting.

Question 3

Find the magnitude of 7i − 24j, and the angle it makes with the positive x-direction, to 1 decimal place.

Accepted 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.

Question 4

A vector has magnitude 10 and makes an angle of 60° with the positive x-direction. Write it in i, j form, giving an exact answer.

Accepted Answer

The components are (10 cos 60°)i + (10 sin 60°)j. M1 for the components in terms of cos and sin. Since cos 60° = 1/2 and sin 60° = √3/2, this is 5i + 5√3j, about 5i + 8.66j. A1 for 5i, A1 for 5√3j.

Question 5

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.

Accepted 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, though its magnitude is unchanged.

Question 6

The vectors 4i + 6j and 6i + λj are parallel. Find the value of λ.

Accepted 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.

Question 7

Find the two vectors of magnitude 26 that are parallel to 5i − 12j.

Accepted 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. Both values are required, because parallel vectors may point in opposite directions. A question asking for the same direction rather than parallel would want only the first.

Question 8

For a = 6i + 8j and b = 8i − 6j, find |a + b|, and show that it is less than |a| + |b|. State the condition on a and b under which |a + b| = |a| + |b|.

Accepted Answer

a + b = 14i + 2j, so |a + b| = √(196 + 4) = √200 = 10√2, about 14.14. M1 for a + b, A1 for 10√2. Here |a| = √(36 + 64) = 10 and |b| = √(64 + 36) = 10, so |a| + |b| = 20, and 10√2 ≈ 14.14 is less than 20. A1 for the comparison. Equality holds when a and b point in exactly the same direction, that is, when one is a positive scalar multiple of the other; here they do not, since neither component pair is in proportion. B1 for the condition. Placed tip to tail, a and b here turn through a right angle rather than continuing in a straight line, which is what makes the sum shorter than the two lengths added.

Question 9

ABCD is a parallelogram with A(3, 2), B(8, 5) and C(11, 10). Find the position vector of D, and verify your answer using the midpoint of each diagonal.

Accepted Answer

In a parallelogram AB = DC, so b − a = c − d, giving d = a + c − b = (3 + 11 − 8, 2 + 10 − 5) = (6, 7). M1 for d = a + c − b, A1 for (6, 7). The midpoint of AC is ((3 + 11)/2, (2 + 10)/2) = (7, 6), and the midpoint of BD is ((8 + 6)/2, (5 + 7)/2) = (7, 6): the two diagonals share a midpoint. M1 for both midpoints, A1 for showing they agree. A parallelogram's diagonals bisect each other, so this check does not depend on which pair of opposite sides was used to find D in the first place.

Question 10

Given p = 2i + j and q = i − j, find the values of s and t such that sp + tq = 7i + 2j.

Accepted Answer

Matching components gives two simultaneous equations: 2s + t = 7 and s − t = 2. M1 for the two equations. Adding them eliminates t: 3s = 9, so s = 3, and then t = s − 2 = 1. A1 for s = 3, A1 for t = 1. Two non-parallel vectors such as p and q reach every point of the plane through exactly one pair of multiples, which is why matching components gives a solvable pair of equations rather than a family of answers.

Question 11

The vector 5i + kj has magnitude 13. Find the possible values of k, and, for the positive value, find the angle the vector makes with the positive x-direction, to 1 decimal place.

Accepted Answer

25 + k² = 169, so k² = 144 and k = ±12. M1 for the magnitude equation, A1 for both values of k. For k = 12, tan θ = 12/5, so θ = 67.4°. M1 for tan θ = 12/5, A1 for the angle.

Question 12

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|.

Accepted 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.

Question 13

The points A(1, 2), B(5, 8) and C(11, 17) are collinear. Show that this is true, and find the ratio AB : BC.

Accepted Answer

AB = b − a = 4i + 6j. M1 for AB. BC = c − b = 6i + 9j. M1 for BC. Since BC = 1.5AB, a scalar multiple, AB and BC are parallel, and A1 for that. They share the point B, so A, B and C lie on one line. B1 for the common point completing the collinearity argument. Because BC = 1.5AB, B divides AC so that AB : BC = 4 : 6 = 2 : 3. A1 for the ratio.

Question 14

Relative to an origin O, points A and B have position vectors a and b. Using vectors, prove that the midpoint M of AB has position vector ½(a + b).

Accepted Answer

AM = ½AB, since M is the midpoint, and AB = b − a, so AM = ½(b − a). M1 for AM = ½(b − a). The position vector of M is OM = OA + AM = a + ½(b − a). M1 for OM = a + AM. Expanding, a + ½(b − a) = a + ½b − ½a = ½a + ½b = ½(a + b). A1 for the simplification. This holds for any two points A and B, since a and b were never given specific values, and it matches the componentwise average used for a midpoint elsewhere in this workbook, because averaging each coordinate is exactly what ½(a + b) does slot by slot. B1 for the argument holding generally, A1 for connecting it to the coordinate average.

Question 15

ABCD is a parallelogram, with position vectors a, b, c, d for its vertices in order. Using the fact that AB and DC are equal as vectors, prove that the diagonals AC and BD of the parallelogram bisect each other.

Accepted Answer

In a parallelogram, opposite sides are equal and parallel, so AB = DC, that is, b − a = c − d. B1 for stating this as the defining property used. Rearranging, d = a + c − b. M1 for the rearrangement. The midpoint of diagonal AC is ½(a + c). M1 for that midpoint. The midpoint of diagonal BD is ½(b + d). M1 for that midpoint. Substituting d: ½(b + d) = ½(b + a + c − b) = ½(a + c). A1 for the substitution and simplification. The two diagonals therefore share the point with position vector ½(a + c), so they bisect each other. A1 for the conclusion.