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Vectors in three dimensions questions

Add a third unit vector, k, and vectors leave the page. Almost nothing else changes: magnitudes still come from Pythagoras, now applied twice, AB is still b minus a, parallel still means scalar multiple, and the geometry of boxes, triangles and midpoints in space reduces to the same slot-by-slot arithmetic.

7 original questions · 24 marks · the vectors in three 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.

• 2 marks · find · warm-up

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

  Worked answer

  |6i − 2j + 3k| = √(36 + 4 + 9) = √49 = 7. M1 for the sum of three squares, A1 for 7. Three squares under one root, which is the two-dimensional formula with one more term. 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.

• 3 marks · warm-up

  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.

  Worked 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.

• 4 marks

  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.

  Worked 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. Averaging −1 and 5 to get 2 is where the negative coordinate usually claims a victim.

• 3 marks

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

  Worked 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. Giving only a = 8 loses the accuracy mark on a question that says values, plural.

• 3 marks · show

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

  Worked 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.

• 6 marks · stretch

  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.

  Worked 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. Trying the relation on AB or AC instead produces a false statement and wastes the working.

• 3 marks · stretch

  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.

  Worked 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.

The same practice on paper: the printable workbook for this topic, questions and a worked answer book.

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