Questions on vectors in three dimensions

Question 1

Find the magnitude of the vector 6i − 2j + 3k.

Accepted Answer

|6i − 2j + 3k| = √(36 + 4 + 9) = √49 = 7. M1 for the sum of three squares, A1 for 7. The formula is Pythagoras applied twice. Run it first across the floor of the box, where the diagonal of the base rectangle has length √(36 + 4) = √40, then again up the vertical, on the right-angled triangle made by that diagonal and the k-component, giving √(40 + 9) = 7.

Question 2

Given a = 6i − 2j + 3k, write down the unit vector in the direction of a, and find a vector of magnitude 21 that is parallel to a.

Accepted Answer

Since |a| = 7, the unit vector is (6/7)i − (2/7)j + (3/7)k. For magnitude 21, scale by 21/7 = 3, giving 18i − 6j + 9k, or its negative −18i + 6j − 9k, since parallel allows either sense. B1 for the unit vector, M1 for scaling by 3, A1 for the vector of magnitude 21. A question asking for the same direction as a would accept only the first of those.

Question 3

The vectors 4i − 6j + 2k and 2i + μj + k are parallel. Find μ.

Accepted Answer

4i − 6j + 2k = 2(2i − 3j + k), so the scalar linking the two vectors is 2. M1 for identifying the scalar from the i-component and the k-component. Matching j-components, 2μ = −6, so μ = −3. A1 for μ = −3.

Question 4

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

Accepted Answer

3p = 6i − 3j + 9k and 2q = −2i + 8j − 4k, so 3p − 2q = (6 − (−2))i + (−3 − 8)j + (9 − (−4))k = 8i − 11j + 13k. M1 for both scalar multiples, A1 for combining them slot by slot, A1 for the printed result.

Question 5

The points A(1, 2, −1) and B(3, −2, 5) have position vectors a and b respectively. Find AB, the exact distance AB, and the midpoint of AB.

Accepted Answer

AB = b − a = 2i − 4j + 6k. The distance is |AB| = √(4 + 16 + 36) = √56 = 2√14, which the word exact requires you to leave in surd form rather than write as 7.48. The midpoint averages the coordinates, giving (2, 0, 2). M1 for b − a, A1 for 2i − 4j + 6k, A1 for 2√14, B1 for the midpoint. The third coordinate of the midpoint is (−1 + 5)/2 = 2.

Question 6

The vector ai + 4j + 8k has magnitude 12. Find the possible values of a.

Accepted Answer

a² + 16 + 64 = 144, so a² = 64 and a = ±8. M1 for the magnitude equation, A1 for a = 8, A1 for a = −8. Both signs survive, because squaring hides the sign, so the answer is two vectors that are mirror images in the plane x = 0.

Question 7

Show that the points P(1, 1, 2), Q(3, 5, 4) and R(6, 11, 7) are collinear.

Accepted Answer

PQ = 2i + 4j + 2k and PR = 5i + 10j + 5k, so PR = 2.5PQ, a scalar multiple. The two vectors are therefore parallel, and they share the point P, so P, Q and R lie on one line. M1 for two of the vectors, A1 for the scalar multiple, B1 for the common point. In three dimensions there is no single cross-multiplication test, so the same multiplier has to work in all three components. Stating that the vectors are parallel without naming the common point leaves the proof incomplete.

Question 8

ABCD is a parallelogram with A(2, 1, 0), B(5, 4, 3) and C(9, 7, 6). Find the position vector of D, and check your answer.

Accepted Answer

In a parallelogram AB = DC, so b − a = c − d, giving d = a + c − b = (2 + 9 − 5, 1 + 7 − 4, 0 + 6 − 3) = (6, 4, 3). M1 for d = a + c − b, A1 for (6, 4, 3). Checking, DC = c − d = (3, 3, 3) and AB = b − a = (3, 3, 3): the two agree, M1 for computing DC, A1 for confirming DC = AB.

Question 9

The points P(1, 2, 1), Q(3, 6, q) and R(9, 18, 13) are collinear. Find q.

Accepted Answer

PQ = 2i + 4j + (q − 1)k and PR = 8i + 16j + 12k. M1 for both vectors. Collinear points make PR a scalar multiple of PQ, and the first two components fix that scalar: 8/2 = 4 and 16/4 = 4, so the multiplier is 4. A1 for the scalar 4. The third component must carry the same multiplier: 12 = 4(q − 1), so q − 1 = 3 and q = 4. M1 for the equation on the third component, A1 for q = 4.

Question 10

Triangle PQR has vertices P(1, 2, 3), Q(5, 4, 7) and R(3, 8, 1). Find the position vector of the midpoint M of QR, and find the vector PM.

Accepted Answer

M averages the position vectors of Q and R: M = ((5 + 3)/2, (4 + 8)/2, (7 + 1)/2) = (4, 6, 4). M1 for the midpoint formula, A1 for (4, 6, 4). Then PM = m − p = (4 − 1, 6 − 2, 4 − 3) = 3i + 4j + k. M1 for PM = m − p, A1 for the printed vector.

Question 11

Find the point on the z-axis that is equidistant from A(1, 2, 3) and B(3, 4, 1).

Accepted Answer

A point on the z-axis has the form (0, 0, z). Its squared distance from A is 1 + 4 + (z − 3)² and from B is 9 + 16 + (z − 1)². M1 for both squared distances in terms of z. Setting them equal: 5 + (z − 3)² = 25 + (z − 1)², so 5 + z² − 6z + 9 = 25 + z² − 2z + 1, giving 14 − 6z = 26 − 2z, so −12 = 4z and z = −3. M1 for expanding and collecting terms, A1 for the equation reduced to one unknown, A1 for z = −3, so the point is (0, 0, −3). Squaring the distances rather than working with the square roots directly keeps the equation in z linear once the squared brackets are expanded.

Question 12

The points A(1, 1, 1), B(2, 3, 3) and C(3, −1, 2) form a triangle. Show that the triangle is right-angled and isosceles.

Accepted Answer

AB = i + 2j + 2k, so |AB| = √(1 + 4 + 4) = 3, and AC = 2i − 2j + k, so |AC| = √(4 + 4 + 1) = 3. Two equal sides makes it isosceles. For the right angle, BC = c − b = i − 4j − k, so |BC|² = 1 + 16 + 1 = 18. Since |AB|² + |AC|² = 9 + 9 = 18 = |BC|², the converse of Pythagoras places the right angle at A, between the two equal sides. M1 for AB and AC, A1 for the two equal lengths, B1 for isosceles, M1 for BC, A1 for |BC|² = 18, A1 for the converse of Pythagoras. Test the longest side as the hypotenuse.

Question 13

The triangle ABC has vertices A(1, 1, 1), B(2, 3, 3) and C(3, −1, 2). Given that the angle at A is a right angle, find the exact area of the triangle.

Accepted Answer

The two sides meeting at the right angle are AB and AC, and they serve as base and perpendicular height. |AB| = √(1 + 4 + 4) = 3 and |AC| = √(4 + 4 + 1) = 3, so the area is ½ × 3 × 3 = 9/2. B1 for using AB and AC as base and height, M1 for ½ × 3 × 3, A1 for 9/2. The right angle is what makes this legitimate. Without it, the product of two side lengths says nothing about the area, and |BC| = √18 is the hypotenuse, so using it as a base would need a perpendicular height that has not been found.

Question 14

The points D(0, 1, 4), E(4, 3, 2) and F(2, 5, 6) form a triangle. Determine, with full working, whether the triangle is isosceles, right-angled, both or neither.

Accepted Answer

DE = 4i + 2j − 2k, so |DE|² = 16 + 4 + 4 = 24. M1 for DE, A1 for |DE|² = 24. DF = 2i + 4j + 2k, so |DF|² = 4 + 16 + 4 = 24, the same as |DE|². M1 for DF, A1 for |DF|² = 24, giving isosceles. EF = −2i + 2j + 4k, so |EF|² = 4 + 4 + 16 = 24 as well, matching the other two sides exactly: the triangle is in fact equilateral. A1 for |EF|² = 24 with the equilateral conclusion. An equilateral triangle has every angle equal to 60°, so it cannot be right-angled; checking the converse of Pythagoras directly confirms this, since 24 + 24 = 48 is not equal to the third side's square, 24. B1 for concluding it is not right-angled, with either reason.

Question 15

A ski lift runs in a straight line from base station A(0, 0, 0) to summit station B(80, 90, 120), with coordinates in metres and the z-axis vertical. Find the length of the lift line, and find the position vector of the point 102 m from A along the line. State one assumption the model makes.

Accepted Answer

AB = 80i + 90j + 120k, so |AB|² = 6400 + 8100 + 14400 = 28900 and |AB| = 170 m. M1 for AB, A1 for 170. The point 102 m from A lies a fraction 102/170 = 3/5 of the way along AB, so its position vector is A + (3/5)AB = (3/5)(80, 90, 120) = (48, 54, 72). M1 for the fraction 3/5, A1 for scaling AB by it, A1 for (48, 54, 72). The model assumes the lift cable runs perfectly straight between the two stations, with no sag and no change of direction along its length. B1 for a stated assumption. Scaling the whole displacement vector AB by a single fraction is what keeps the point on the line; scaling each coordinate of B by 3/5 instead would give a different point unless A were the origin, which happens to be true here but is not true in general.