Pure: Vectors
Vectors are quantities that have both magnitude and direction, and A Level Maths uses them to describe position, displacement and forces in two and three dimensions with i, j and k components. Pure vector questions cover magnitude, dividing a line in a given ratio, midpoints, and using vectors to prove points are collinear or a shape is a parallelogram.
Before you start
Make sure you're comfortable with these topics first:
Method
- Write a vector between two points as (position vector of the end point) minus (position vector of the start point), e.g. AB = OB - OA.
- Find the magnitude of a vector ai + bj (+ ck) using Pythagoras: |v| = sqrt(a^2 + b^2 (+ c^2)).
- To find a unit vector in the direction of v, divide v by its magnitude |v|.
- To find a point that divides a line segment in a given ratio, use the section formula: if R divides PQ so that PR : RQ = m : n, then OR = OP + (m/(m+n))(OQ - OP).
- To show three points are collinear, find two vectors between pairs of the points and show one is a scalar multiple of the other, with a common point linking them.
- To prove a shape is a parallelogram, show one pair of opposite sides are equal and parallel (i.e. represented by the same vector).
Worked example
Relative to a fixed origin O, points A and B have position vectors OA = 2i - j + 3k and OB = 6i + 3j - 5k. Find the vector AB, its magnitude as an exact simplified surd, and the position vector of the midpoint M of AB.
- AB = OB - OA = (6i + 3j - 5k) - (2i - j + 3k) = 4i + 4j - 8k.
- |AB| = sqrt(4^2 + 4^2 + (-8)^2) = sqrt(16 + 16 + 64) = sqrt(96) = 4sqrt(6).
- M = (OA + OB)/2 = ((2+6)i + (-1+3)j + (3-5)k)/2 = (8i + 2j - 2k)/2.
- So the midpoint has position vector OM = 4i + j - k.
- Final answer: AB = 4i + 4j - 8k, |AB| = 4sqrt(6), and OM = 4i + j - k.
Practice questions
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Q1The vector a = 5i - 12j. Find |a|.Show answer
Answer: 13 (sqrt(5^2 + 12^2) = sqrt(169)).
Q2Given b = 3i + 4j, find the unit vector in the direction of b.Show answer
Answer: (3/5)i + (4/5)j (divide b by |b| = 5).
Q3Relative to a fixed origin O, OA = i + 2j - 2k and OB = 5i - 2j + 4k. Find the vector AB.Show answer
Answer: AB = 4i - 4j + 6k (OB - OA).
Q4Points P and Q have position vectors p = 2i + j and q = 8i + 7j. Find the position vector of R, where R lies on PQ with PR : RQ = 1 : 2.Show answer
Answer: r = 4i + 3j (r = p + (1/3)(q - p)).
Q5OABC is a quadrilateral with position vectors 0, a, a+c and c relative to O, where a and c are non-zero, non-parallel vectors. Show that OABC is a parallelogram.Show answer
Answer: AB = OB - OA = c = OC, so one pair of opposite sides (AB and OC) is equal and parallel, so OABC is a parallelogram.
Q6Relative to a fixed origin O, OA = 3i - j + k, OB = 7i + 3j - 3k and OC = (4t+3)i + 7j - 7k, where t is a constant. Given that A, B and C are collinear, find the value of t.Show answer
Answer: t = 2 (find AB = 4i+4j-4k and AC = 4ti+8j-8k; the j- and k-components show AC = 2AB, so 4t = 8).
Exam-style questions
Written in the style of a A Level Maths exam paper, with a full mark scheme.
Relative to a fixed origin O, the points C and D have position vectors OC = 3i + 5j - 2k and OD = -i + 9j + 6k. (a) Find the vector CD. (2) (b) Find |CD|, giving your answer as an exact simplified surd. (2)
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Three forces acting on a particle are F1 = (4i - 3j) N, F2 = (-7i + 2j) N and F3 = (pi + qj) N. The particle is in equilibrium. (a) Find the values of p and q. (3) (b) Find the magnitude of F3, giving your answer in newtons to 3 significant figures. (2)
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In triangle OAB, OA = a and OB = b. The point P lies on OA such that OP = (3/5)a, and the point Q lies on AB such that AQ = (1/3)AB. (a) Find AB and OQ in terms of a and b. (3) (b) Show that PQ = (1/15)(a + 5b), fully simplified. (3)
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