Salam Academy
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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14 questions

SA-9709-P3-VE-0031 markL1 FoundationCalculatorScalar product
Find a⋅b\mathbf{a}\cdot\mathbf{b} where a=(3−12)\mathbf{a}=\begin{pmatrix}3\\-1\\2\end{pmatrix} and b=(14−2)\mathbf{b}=\begin{pmatrix}1\\4\\-2\end{pmatrix}.

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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-VE2-0051 markL1 FoundationCalculatorScalar product
∣a∣=3|\mathbf{a}|=3, ∣b∣=4|\mathbf{b}|=4 and the angle between a\mathbf{a} and b\mathbf{b} is 60∘60^\circ. Find a⋅b\mathbf{a}\cdot\mathbf{b}.

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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-VE2-G011 markL1 FoundationCalculatorScalar product
a=2i+j+3k\mathbf{a}=2\mathbf{i}+\mathbf{j}+3\mathbf{k} and b=i−4j+2k\mathbf{b}=\mathbf{i}-4\mathbf{j}+2\mathbf{k}. Find a⋅b\mathbf{a}\cdot\mathbf{b}.

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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-VE2-Q011 markL1 FoundationCalculatorScalar product
a⋅b<0\mathbf{a}\cdot\mathbf{b}<0 tells you that the angle between a\mathbf{a} and b\mathbf{b} is
Choose one option

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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-VE-Q031 markL1 FoundationCalculatorPerpendicular
For non-zero vectors, a⋅b=0\mathbf{a}\cdot\mathbf{b}=0 means
Choose one option

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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-VE-0052 marksL2 StandardCalculatorPerpendicular
The vectors (2p−1)\begin{pmatrix}2\\p\\-1\end{pmatrix} and (314)\begin{pmatrix}3\\1\\4\end{pmatrix} are perpendicular. Find pp.

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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-VE2-0013 marksL2 StandardCalculatorScalar product
Find the angle between a=3i+4k\mathbf{a}=3\mathbf{i}+4\mathbf{k} and b=5j+12k\mathbf{b}=5\mathbf{j}+12\mathbf{k}, in degrees correct to 1 decimal place.

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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-VE2-0022 marksL2 StandardCalculatorPerpendicular
The vectors (k2−1)\begin{pmatrix}k\\2\\-1\end{pmatrix} and (3k4)\begin{pmatrix}3\\k\\4\end{pmatrix} are perpendicular. Find kk.

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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-VE2-0043 marksL2 StandardCalculatorPerpendicular
A(1,0,2)A(1,0,2), B(3,1,1)B(3,1,1) and C(4,k,3)C(4,k,3). Angle ABCABC is a right angle. Find kk.

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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-VE2-0072 marksL2 StandardCalculatorScalar product
Find the angle between the vector 2i+3j+6k2\mathbf{i}+3\mathbf{j}+6\mathbf{k} and the xx-axis, in degrees correct to 1 decimal place.

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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-VE-0043 marksL3 ExamCalculatorScalar product
Find the angle between p=i+2j+2k\mathbf{p}=\mathbf{i}+2\mathbf{j}+2\mathbf{k} and q=2i−j+2k\mathbf{q}=2\mathbf{i}-\mathbf{j}+2\mathbf{k}, giving your answer in degrees correct to 1 decimal place.

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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-VE2-0034 marksL3 ExamCalculatorTriangles
The points A(1,2,0)A(1,2,0), B(3,3,2)B(3,3,2) and C(2,0,3)C(2,0,3) form a triangle. Find angle BACBAC, in degrees correct to 1 decimal place.

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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-VE2-0064 marksL3 ExamCalculatorPerpendicular
The vector i+pj+qk\mathbf{i}+p\mathbf{j}+q\mathbf{k} is perpendicular to both i+j+k\mathbf{i}+\mathbf{j}+\mathbf{k} and i−j+2k\mathbf{i}-\mathbf{j}+2\mathbf{k}. Find pp and qq.

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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-VE2-0085 marksL4 ChallengingCalculatorTriangles
OA→=i+2j+2k\overrightarrow{OA}=\mathbf{i}+2\mathbf{j}+2\mathbf{k} and OB→=2i−j+2k\overrightarrow{OB}=2\mathbf{i}-\mathbf{j}+2\mathbf{k}. Find the exact area of triangle OABOAB.

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Salam Academy Original Practice · Proposed mark scheme