Edexcel A level Maths (9MA0) · Pure mathematics › Vectors
Practise Vector arithmetic and geometry. 4 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy5 marks
(a) The vectors \(\mathbf{a} = -2\mathbf{i} - 6\mathbf{j}\) and \(\mathbf{b} = p\mathbf{i} - 24\mathbf{j}\), where \(p\) is a constant, are parallel. Find the value of \(p\).[2]
(b) The vectors \(\mathbf{c} = -2\mathbf{i} + \mathbf{j} + q\mathbf{k}\) and \(\mathbf{d} = r\mathbf{i} - 3\mathbf{j} - 15\mathbf{k}\), where \(q\) and \(r\) are constants, are parallel. Find the values of \(q\) and \(r\).[3]
Show the answer and mark scheme
(a)Answer: \(p = -8\)
M1 for \(\mathbf{b} = k\mathbf{a}\) with \(k = 4\) found from the known components (or equal ratios)
A1 for \(p = -8\)
Worked solution: Comparing the \(\mathbf{j}\) components: \(-24 = 4 \times (-6)\), so \(\mathbf{b} = 4\mathbf{a}\). \(p = 4 \times (-2) = -8\)
(b)Answer: \(q = 5, \ r = 6\)
M1 for \(\mathbf{d} = k\mathbf{c}\) with \(k = -3\) from the \(\mathbf{j}\) components
A1 for \(q = 5\)
A1 for \(r = 6\)
Worked solution: \(\mathbf{j}\): \(-3 = k \times 1 \Rightarrow k = -3\) \(\mathbf{k}\): \(-15 = -3q \Rightarrow q = 5\) \(\mathbf{i}\): \(r = -3 \times (-2) = 6\)
Question 2Medium6 marks
The diagram shows a triangle \(OAB\) with \(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\). The points \(M\) and \(N\) are the midpoints of \(OA\) and \(OB\) respectively.
[object Object]
(a) Prove, using vectors, that \(MN\) is parallel to \(AB\) and that \(MN = \tfrac{1}{2}AB\).[4]
(b) Hence show that the area of triangle \(OMN\) is one quarter of the area of triangle \(OAB\).[2]
Show the answer and mark scheme
(a)Answer: \(\overrightarrow{MN} = \tfrac{1}{2}\mathbf{b} - \tfrac{1}{2}\mathbf{a} = \tfrac{1}{2}(\mathbf{b} - \mathbf{a}) = \tfrac{1}{2}\overrightarrow{AB}\), so \(MN \parallel AB\) and \(MN = \tfrac{1}{2}AB\).
M1 for \(\overrightarrow{OM} = \tfrac{1}{2}\mathbf{a}\) and \(\overrightarrow{ON} = \tfrac{1}{2}\mathbf{b}\)
A1 for \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\) and \(\overrightarrow{MN} = \tfrac{1}{2}\overrightarrow{AB}\)
A1* for a complete conclusion: \(\overrightarrow{MN}\) is a scalar multiple of \(\overrightarrow{AB}\), so the lines are parallel, and the length of \(MN\) is half the length of \(AB\)
Worked solution: \(\overrightarrow{MN} = \overrightarrow{MO} + \overrightarrow{ON} = -\tfrac{1}{2}\mathbf{a} + \tfrac{1}{2}\mathbf{b} = \tfrac{1}{2}(\mathbf{b} - \mathbf{a})\) \(\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} = \mathbf{b} - \mathbf{a}\), so \(\overrightarrow{MN} = \tfrac{1}{2}\overrightarrow{AB}\). Hence \(MN\) is parallel to \(AB\) and \(MN = \tfrac{1}{2}AB\).
(b)Answer: Triangles \(OMN\) and \(OAB\) are similar with scale factor \(\tfrac{1}{2}\), so the ratio of areas is \(\left(\tfrac{1}{2}\right)^2 = \tfrac{1}{4}\).
M1 for identifying that the triangles are similar (corresponding sides in the ratio \(1 : 2\), or angle \(O\) common and \(MN \parallel AB\))
A1* for area scale factor \(\left(\tfrac{1}{2}\right)^2 = \tfrac{1}{4}\), or \(\tfrac{1}{2} \times \tfrac{1}{2}OA \times \tfrac{1}{2}OB \sin O = \tfrac{1}{4} \times \tfrac{1}{2}OA \times OB\sin O\)
Worked solution: \(OM = \tfrac{1}{2}OA\), \(ON = \tfrac{1}{2}OB\) and angle \(MON\) = angle \(AOB\). \(\text{Area } OMN = \tfrac{1}{2} \times \tfrac{1}{2}OA \times \tfrac{1}{2}OB \times \sin AOB = \tfrac{1}{4}\left(\tfrac{1}{2} \times OA \times OB \times \sin AOB\right) = \tfrac{1}{4}\,\text{Area } OAB\)
Question 3Hard7 marks
The diagram shows a parallelogram \(OABC\) with \(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OC} = \mathbf{c}\). The diagonals \(OB\) and \(AC\) intersect at the point \(X\).
[object Object]
(a) Given that \(\overrightarrow{OX} = \lambda\overrightarrow{OB}\) and \(\overrightarrow{AX} = \mu\overrightarrow{AC}\), find two expressions for \(\overrightarrow{OX}\) in terms of \(\lambda\), \(\mu\), \(\mathbf{a}\) and \(\mathbf{c}\).[3]
(b) Hence prove that the diagonals of the parallelogram bisect each other.[3]
(c) Explain why it is valid to compare the coefficients of \(\mathbf{a}\) and \(\mathbf{c}\) in part (b).[1]
(b)Answer: Comparing coefficients: \(\lambda = 1 - \mu\) and \(\lambda = \mu\), so \(\lambda = \mu = \tfrac{1}{2}\): \(X\) is the midpoint of both diagonals.
M1 for equating the coefficients of \(\mathbf{a}\) and of \(\mathbf{c}\): \(\lambda = 1 - \mu, \ \lambda = \mu\)
A1 for \(\lambda = \mu = \tfrac{1}{2}\)
A1* for the conclusion that \(X\) is the midpoint of \(OB\) and of \(AC\), so the diagonals bisect each other
Worked solution: \(\lambda\mathbf{a} + \lambda\mathbf{c} = (1 - \mu)\mathbf{a} + \mu\mathbf{c}\). Since \(\mathbf{a}\) and \(\mathbf{c}\) are not parallel, compare coefficients: \(\lambda = 1 - \mu\) and \(\lambda = \mu\), giving \(\lambda = \mu = \tfrac{1}{2}\). So \(\overrightarrow{OX} = \tfrac{1}{2}\overrightarrow{OB}\) and \(\overrightarrow{AX} = \tfrac{1}{2}\overrightarrow{AC}\): \(X\) is the midpoint of each diagonal.
(c)Answer: \(\mathbf{a}\) and \(\mathbf{c}\) are non-zero and not parallel, so \(p\mathbf{a} + q\mathbf{c} = r\mathbf{a} + s\mathbf{c}\) implies \(p = r\) and \(q = s\).
B1 for \(\mathbf{a}\) and \(\mathbf{c}\) are non-zero and not parallel (so neither can be written as a multiple of the other)
Worked solution: If \(p\mathbf{a} + q\mathbf{c} = r\mathbf{a} + s\mathbf{c}\) with \(p \ne r\), then \(\mathbf{a} = \frac{s - q}{p - r}\mathbf{c}\), making \(\mathbf{a}\) parallel to \(\mathbf{c}\), which is false. So the coefficients must be equal.