Worksheets · Higher

Vector geometric arguments and proofs

8 exam-style questions, grades 6 to 9. Worked solutions and the marks 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 · 2 marks
(a) OA = a and OB = b are position vectors. M is the midpoint of AB. P lies on OB with OP:PB = 2:3. Express the vector PM in terms of a and b. (2)
Question 3Non-calculator · 5 marks
OAB is a triangle with OA = a and OB = b. M is the midpoint of OA and N lies on AB with AN:NB = 1:2.
(a) Express MN in terms of a and b. (3)
(b) P lies on OB with OP:PB = 2:1. A pupil claims MN is parallel to AP. Decide whether the claim is true, giving a vector argument. (2)
Question 4Non-calculator · 4 marks
In triangle OAB, OA = a and OB = b. M and N divide OA and OB in the same ratio OM:MA = ON:NB = 2:3.
(a) Express MN in terms of a and b. (2)
(b) Prove MN is parallel to AB and state MN:AB. (2)
Question 5Non-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 6Non-calculator · 4 marks
$\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$ . $P$ is the point on $AB$ such that $AP : PB = 2 : 1$ . $Q$ is the point with $\overrightarrow{OQ} = \mathbf{a} + 2\mathbf{b}$ .
(a) Prove that $O$ , $P$ and $Q$ lie on a straight line. (4)
Question 7Non-calculator · 4 marks
(a) OAB is a triangle. M lies on OA with OM:MA = 1:2. N lies on AB with AN:NB = 2:1. Lines BM and ON intersect at X. Work out the fraction OX/ON. Show your working using vectors. (4)
Question 8Non-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)

Worked solutions and marks

Question 1

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

$\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} = -\mathbf{a} + \mathbf{b}$ .

B1 $\mathbf{b} - \mathbf{a}$ .

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

$\overrightarrow{OM} = \overrightarrow{OA} + \tfrac{1}{2}\overrightarrow{AB} = \mathbf{a} + \tfrac{1}{2}(\mathbf{b} - \mathbf{a})$
$= \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}$ .

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

Question 2

(a) $1/2 a + 1/10 b$

$(a+b)/2-2b/5$
The midpoint has position vector OM = (a + b)/2.
OP = (2/5)b because OP is two of the five equal parts of OB.
PM = OM $-$ OP = (1/2)a + (1/2 $-$ 2/5)b = (1/2)a + (1/10)b.

P1 Establishing $(a+b)/2-2b/5$ or an equivalent valid method.
A1 Correct answer: $1/2 a + 1/10 b$

Question 3

(a) $a/6+b/3$

Find N from A plus one third of AB.
$a+(b-a)/3$
Subtract M’s position vector.
$2a/3+b/3-a/2$
Therefore $a/6+b/3$ .

M1 Find N from A plus one third of AB.
M1 Subtract M’s position vector.
A1 Correct answer: $a/6+b/3$

(b) AP = 2b/3 - a. This is not a scalar multiple of MN = a/6+b/3; the stated parallel claim is false.

Compare the two independent vector coefficients.
Therefore AP = 2b/3 - a. This is not a scalar multiple of MN = a/6+b/3; the stated parallel claim is false.

M1 Compare the two independent vector coefficients.
C1 Correct conclusion with supporting reasoning: AP = 2b/3 - a. This is not a scalar multiple of MN = a/6+b/3; the stated parallel claim is false.

Question 4

(a) $2(b-a)/5$

Subtract OM from ON.
$2b/5-2a/5$
Therefore $2(b-a)/5$ .

M1 Subtract OM from ON.
A1 Correct answer: $2(b-a)/5$

(b) MN = (2/5)(b-a) = (2/5)AB, so the vectors are parallel in the same direction and the length ratio is 2:5.

Compare with the displacement AB = b-a.
$2(b-a)/5$
Therefore MN = (2/5)(b-a) = (2/5)AB, so the vectors are parallel in the same direction and the length ratio is 2:5.

M1 Compare with the displacement AB = b-a.
C1 Correct conclusion with supporting reasoning: MN = (2/5)(b-a) = (2/5)AB, so the vectors are parallel in the same direction and the length ratio is 2:5.

Question 5

(a) $2a/3+b$

Move from A to C, then one third of CD.
$a+b-a/3$
Therefore $2a/3+b$ .

M1 Move from A to C, then one third of CD.
A1 Correct answer: $2a/3+b$

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

M has position a+b/2; subtract it from AN.
$(2a/3+b)-(a+b/2)$
Therefore $-a/3+b/2$ .

M1 M has position a+b/2; subtract it from AN.
A1 Correct answer: $-a/3+b/2$

Question 6

(a) $\overrightarrow{OP} = \frac{1}{3}(\mathbf{a} + 2\mathbf{b})$ , so $\overrightarrow{OQ} = 3\overrightarrow{OP}$ .

$\overrightarrow{OP} = \mathbf{a} + \tfrac{2}{3}(\mathbf{b} - \mathbf{a}) = \tfrac{1}{3}\mathbf{a} + \tfrac{2}{3}\mathbf{b} = \tfrac{1}{3}(\mathbf{a} + 2\mathbf{b})$
So $\overrightarrow{OQ} = 3\overrightarrow{OP}$ .
$\overrightarrow{OP}$ and $\overrightarrow{OQ}$ are parallel and share the point $O$ , so $O$ , $P$ and $Q$ are collinear.

M1 $\overrightarrow{AP} = \frac{2}{3}(\mathbf{b} - \mathbf{a})$ .
M1 $\overrightarrow{OP} = \frac{1}{3}\mathbf{a} + \frac{2}{3}\mathbf{b}$ .
M1 Showing $\overrightarrow{OQ} = 3\overrightarrow{OP}$ (a scalar multiple).
C1 Concluding collinear because the vectors are parallel and share the point $O$ .

Question 7

(a) $\frac{3}{5}$

Let OA = a and OB = b. Then OM = a/3 and ON = a + (2/3)(b $-$ a) = a/3 + 2b/3.
Write OX = t ON = (t/3)a + (2t/3)b. Also, because X lies on BM, OX = b + s(a/3 $-$ b) = (s/3)a + (1 $-$ s)b.
The independent vector coefficients give t = s and 2t/3 = 1 $-$ s.
Let OA = a and OB = b. Then OM = a/3 and ON = a + (2/3)(b $-$ a) = a/3 + 2b/3.
Write OX = t ON = (t/3)a + (2t/3)b. Also, because X lies on BM, OX = b + s(a/3 $-$ b) = (s/3)a + (1 $-$ s)b.
The independent vector coefficients give t = s and 2t/3 = 1 $-$ s.
Therefore 2t/3 = 1 $-$ t, so 5t/3 = 1 and OX/ON = t = 3/5.

P1 Let OA = a and OB = b. Then OM = a/3 and ON = a + (2/3)(b $-$ a) = a/3 + 2b/3.
P1 Write OX = t ON = (t/3)a + (2t/3)b. Also, because X lies on BM, OX = b + s(a/3 $-$ b) = (s/3)a + (1 $-$ s)b.
P1 The independent vector coefficients give t = s and 2t/3 = 1 $-$ s.
A1 Correct answer: $\frac{3}{5}$

Question 8

(a) $-1$

Take A as the origin and write AB = b and AC = c. Then AM = b/2 and AN = c/3.
Since P lies on MN, AP = b/2 + t(c/3 $-$ b/2) = ((1 $-$ t)/2)b + (t/3)c.
Since BP = k BC, AP = b + k(c $-$ b) = (1 $-$ k)b + kc.
Take A as the origin and write AB = b and AC = c. Then AM = b/2 and AN = c/3.
Since P lies on MN, AP = b/2 + t(c/3 $-$ b/2) = ((1 $-$ t)/2)b + (t/3)c.
Since BP = k BC, AP = b + k(c $-$ b) = (1 $-$ k)b + kc.
Equating coefficients gives t = 3k and (1 $-$ t)/2 = 1 $-$ k. Hence 1 $-$ 3k = 2 $-$ 2k.
Therefore k = $-1.$ P lies beyond B, so BP has the opposite direction to BC.

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: $-1$

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Vector geometric arguments and proofs

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