Vector geometric arguments and proofs

3 exam-style questions. Answers and the mark for each step are on the last page.

Question 1Non-calculator · 3 marks
$\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$ . $M$ is the midpoint of $AB$ .
(a) Find $\overrightarrow{AB}$ in terms of $\mathbf{a}$ and $\mathbf{b}$ . (1)
(b) Find $\overrightarrow{OM}$ in terms of $\mathbf{a}$ and $\mathbf{b}$ . Give your answer in its simplest form. (2)
Question 2Non-calculator · 4 marks
ABCD is a parallelogram with AB = a and AD = b. M is the midpoint of BC. N lies on CD with CN:ND = 1:2.
(a) Express AN in terms of a and b. (2)
(b) Express MN in terms of a and b. (2)
Question 3Non-calculator · 4 marks
(a) ABC is a triangle. M is the midpoint of AB. N lies on AC with AN:NC = 1:2. The line MN meets the line BC extended at P. The vector BP equals k times the vector BC. Work out k. Show your working using vectors. (4)

Answers and marks

(a) $\mathbf{b} - \mathbf{a}$

(b) $\frac{1}{2}(\mathbf{a} + \mathbf{b})$

M1 $\mathbf{a} + \frac{1}{2}(\mathbf{b} - \mathbf{a})$ or $\mathbf{b} + \frac{1}{2}(\mathbf{a} - \mathbf{b})$ .
A1 $\frac{1}{2}(\mathbf{a} + \mathbf{b})$ .

(a) $2a/3+b$

(b) $-a/3+b/2$

(a) $-1$

P1 Take A as the origin and write AB = b and AC = c. Then AM = b/2 and AN = c/3.
P1 Since P lies on MN, AP = b/2 + t(c/3 $-$ b/2) = ((1 $-$ t)/2)b + (t/3)c.
P1 Since BP = k BC, AP = b + k(c $-$ b) = (1 $-$ k)b + kc.
A1 Correct answer: $-1$

Vector geometric arguments and proofs

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