Edexcel A level Maths (9MA0) · Pure mathematics › Vectors
Practise Position vectors and distances. unlimited generated questions 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 1Easy6 marks
Relative to a fixed origin \(O\), the points \(A\) and \(B\) have position vectors\[\overrightarrow{OA} = -4\mathbf{i} + 4\mathbf{j}, \qquad \overrightarrow{OB} = -6\mathbf{i} - 6\mathbf{j}\]respectively.
(a) Find \(\overrightarrow{AB}\).[2]
(b) Find the exact distance \(AB\), giving your answer as a simplified surd where appropriate.[2]
(c) Find the position vector of the midpoint of \(AB\).[2]
M1 for \(\overrightarrow{OD} = \overrightarrow{OA} + \overrightarrow{BC}\) or \(\overrightarrow{OC} + \overrightarrow{BA}\)
A1 for \(3\mathbf{i} - 5\mathbf{k}\)
Worked solution: In the parallelogram \(ABCD\), \(\overrightarrow{AD} = \overrightarrow{BC} = 4\mathbf{i} - 4\mathbf{j}\), so \(\overrightarrow{OD} = \overrightarrow{OA} + \overrightarrow{BC} = 3\mathbf{i} - 5\mathbf{k}\)
(c)Answer: \(2\sqrt{6}\)
M1 for the midpoint \(N\) of \(BC\) (\(\overrightarrow{ON} = -2\mathbf{i} + 3\mathbf{j} - 6\mathbf{k}\)) or for \(AN^2 = AB^2 - \left(\tfrac{1}{2}BC\right)^2\)
M1 for \(AN = \sqrt{3}\) (the perpendicular height, since the triangle is isosceles)
A1 for \(2\sqrt{6}\) oe
Worked solution: As \(AB = AC\), the line from \(A\) to the midpoint \(N\) of \(BC\) is perpendicular to \(BC\). \(AN^2 = AB^2 - \left(\tfrac{1}{2}BC\right)^2 = 11 - \frac{32}{4} = 3\) \(\text{Area} = \tfrac{1}{2} \times BC \times AN = \tfrac{1}{2} \times 4\sqrt{2} \times \sqrt{3} = 2\sqrt{6}\)
Question 3Hard10 marks
Relative to a fixed origin \(O\), the points \(A\), \(B\) and \(C\) have position vectors\[\overrightarrow{OA} = 5\mathbf{i} - 3\mathbf{k}, \qquad \overrightarrow{OB} = \mathbf{i} + 2\mathbf{j} - \mathbf{k}, \qquad \overrightarrow{OC} = 5\mathbf{i} + 3\mathbf{j} + \mathbf{k}\]respectively.
(a) Find the exact lengths of \(AB\), \(BC\) and \(AC\).[3]
(b) Find the size of angle \(ABC\). Give your answer in degrees to one decimal place.[3]
(c) Find the area of triangle \(ABC\), giving your answer to 3 significant figures.[2]
(d) Hence find the shortest distance from \(A\) to the line through \(B\) and \(C\), giving your answer to 3 significant figures.[2]
Show the answer and mark scheme
(a)Answer: \(AB = 2\sqrt{6}, \ BC = \sqrt{21}, \ AC = 5\)
M1 for a correct method for one length, e.g. \(AB = \sqrt{(-4)^2 + 2^2 + 2^2}\)
M1 for \(\tfrac{1}{2} \times 2\sqrt{6} \times \sqrt{21} \times \sin ABC\) with their angle
A1 for 10.0 (awrt)
Worked solution: \(\text{Area} = \tfrac{1}{2} \times AB \times BC \times \sin ABC = \tfrac{1}{2} \times 2\sqrt{6} \times \sqrt{21} \times \sin 63.5^\circ = 10.0\) (3 s.f.)
(d)Answer: \(4.39\)
M1 for \(\tfrac{1}{2} \times BC \times h = \text{Area}\) or \(h = AB\sin ABC\)
A1 for 4.39 (awrt)
Worked solution: The shortest distance is the perpendicular height from \(A\) to \(BC\): \(h = \frac{2 \times \text{Area}}{BC} = \frac{2 \times 10.05}{\sqrt{21}} = 4.39\) (3 s.f.)