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Practise

Question bank

Original exam-style questions, mapped to your syllabus. Answers are checked for mathematical equivalence and every attempt updates your mastery.

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12 questions

SA-4024-TV-0111 markL1 FoundationCalculatorVector paths
OA→=a\overrightarrow{OA}=\mathbf a, OB→=b\overrightarrow{OB}=\mathbf b and MM is the midpoint of ABAB. Then OM→=\overrightarrow{OM}=
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Salam Academy Original Practice · Proposed mark scheme
SA-4024-TV-0121 markL1 FoundationCalculatorVector paths
OA→=a\overrightarrow{OA}=\mathbf a and OB→=b\overrightarrow{OB}=\mathbf b. Then AB→=\overrightarrow{AB}=
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Salam Academy Original Practice · Proposed mark scheme
SA-4024-TV4-0011 markL1 FoundationCalculatorVector paths
OA→=a\overrightarrow{OA}=\mathbf a and OB→=b\overrightarrow{OB}=\mathbf b. Then BA→=\overrightarrow{BA}=
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Salam Academy Original Practice · Proposed mark scheme
SA-4024-TV4-0031 markL1 FoundationCalculatorVector paths
OABCOABC is a parallelogram with OA→=a\overrightarrow{OA}=\mathbf a and OC→=c\overrightarrow{OC}=\mathbf c. Then OB→=\overrightarrow{OB}=
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Salam Academy Original Practice · Proposed mark scheme
SA-4024-TV4-0051 markL1 FoundationNon-calculatorParallel & collinear
(4−6)=k(−23)\dbinom4{-6}=k\dbinom{-2}3. Find kk.

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Salam Academy Original Practice · Proposed mark scheme
SA-4024-TV4-0071 markL1 FoundationCalculatorParallel & collinear
AB→=(23)\overrightarrow{AB}=\dbinom23 and AC→=(69)\overrightarrow{AC}=\dbinom69. Which statement is true?
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Salam Academy Original Practice · Proposed mark scheme
SA-4024-TV-0132 marksL2 StandardNon-calculatorParallel & collinear
p=(23)\mathbf p=\dbinom23 and q=(69)\mathbf q=\dbinom69. Find kk such that q=kp\mathbf q=k\mathbf p, and state what this shows.
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Salam Academy Original Practice · Proposed mark scheme
SA-4024-TV-0142 marksL2 StandardNon-calculatorParallel & collinear
The vector (k6)\dbinom{k}{6} is parallel to (23)\dbinom23. Find kk.

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Salam Academy Original Practice · Proposed mark scheme
SA-4024-TV4-0021 markL2 StandardCalculatorVector paths
OA→=a\overrightarrow{OA}=\mathbf a and OB→=b\overrightarrow{OB}=\mathbf b. MM is the midpoint of OAOA. Then BM→=\overrightarrow{BM}=
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Salam Academy Original Practice · Proposed mark scheme
SA-4024-TV4-0043 marksL2 StandardCalculatorVector paths
OA→=a\overrightarrow{OA}=\mathbf a and OB→=b\overrightarrow{OB}=\mathbf b. PP lies on ABAB with AP:PB=3:1AP:PB=3:1. Write OP→=λa+μb\overrightarrow{OP}=\lambda\mathbf a+\mu\mathbf b and find λ\lambda and μ\mu.

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Salam Academy Original Practice · Proposed mark scheme
SA-4024-TV4-0062 marksL2 StandardCalculatorParallel & collinear
AB→=2a+b\overrightarrow{AB}=2\mathbf a+\mathbf b and BC→=6a+3b\overrightarrow{BC}=6\mathbf a+3\mathbf b. (i) Find kk such that BC→=kAB→\overrightarrow{BC}=k\overrightarrow{AB}. (ii) What does this show about AA, BB and CC?
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Salam Academy Original Practice · Proposed mark scheme
SA-4024-TV-0165 marksL4 ChallengingCalculatorVector paths
OA→=a\overrightarrow{OA}=\mathbf a and OB→=b\overrightarrow{OB}=\mathbf b. PP lies on ABAB with AP:PB=1:2AP:PB=1:2. (i) Write OP→=λa+μb\overrightarrow{OP}=\lambda\mathbf a+\mu\mathbf b and state λ\lambda and μ\mu. (ii) Given a=(30)\mathbf a=\dbinom30 and b=(06)\mathbf b=\dbinom06, find ∣OP→∣|\overrightarrow{OP}|, to 3 significant figures.

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Salam Academy Original Practice · Proposed mark scheme