Edexcel A level Maths (9MA0) · Pure mathematics › Vectors
Practise Vector problems. 1 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
[In this question \(\mathbf{i}\) and \(\mathbf{j}\) are perpendicular unit vectors in a horizontal plane.] Two forces \(\mathbf{F}_1 = (-6\mathbf{i} + \mathbf{j})\text{ N}\) and \(\mathbf{F}_2 = (-6\mathbf{i} + 5\mathbf{j})\text{ N}\) act on a particle \(P\).
(a) Find the resultant force \(\mathbf{R}\) acting on \(P\).[1]
(b) Find the magnitude of \(\mathbf{R}\), giving your answer to 3 significant figures.[2]
(c) Find the angle between the direction of \(\mathbf{R}\) and the vector \(\mathbf{i}\). Give your answer in degrees to one decimal place.[2]
Worked solution: \(|\mathbf{R}| = \sqrt{(-12)^2 + 6^2} = 6\sqrt{5} = 13.4\) N (3 s.f.)
(c)Answer: \(153.4^\circ\) °
M1 for \(\tan\theta = \frac{6}{12}\) or \(\cos\theta = \frac{-12}{|\mathbf{R}|}\) oe
A1 for \(153.4^\circ\)
Worked solution: \(\tan^{-1}\frac{6}{12} = 26.6^\circ\); \(\mathbf{R}\) has a negative \(\mathbf{i}\) component, so the angle with \(\mathbf{i}\) is \(180^\circ - 26.6^\circ = 153.4^\circ\)
Question 2Medium6 marks
[In this question \(\mathbf{i}\) and \(\mathbf{j}\) are perpendicular unit vectors in a horizontal plane.] Three forces \(\mathbf{F}_1\), \(\mathbf{F}_2\) and \(\mathbf{F}_3\) act on a particle \(P\), where \(\mathbf{F}_1 = (3\mathbf{i} + 2\mathbf{j})\text{ N}\) and \(\mathbf{F}_2 = (6\mathbf{i} + 4\mathbf{j})\text{ N}\). The particle is in equilibrium.
(a) Find \(\mathbf{F}_3\).[2]
(b) Find the magnitude of \(\mathbf{F}_3\), giving your answer to 3 significant figures.[2]
(c) Find the angle between the direction of \(\mathbf{F}_3\) and the vector \(\mathbf{j}\). Give your answer in degrees to one decimal place.[2]
M1 for \(\mathbf{F}_1 + \mathbf{F}_2 + \mathbf{F}_3 = \mathbf{0}\)
A1 for \((-9\mathbf{i} - 6\mathbf{j})\text{ N}\)
Worked solution: In equilibrium the resultant force is zero: \(\mathbf{F}_3 = -(\mathbf{F}_1 + \mathbf{F}_2) = -(9\mathbf{i} + 6\mathbf{j}) = (-9\mathbf{i} - 6\mathbf{j})\text{ N}\)
(b)Answer: \(3\sqrt{13} = 10.8\) N
M1 for \(\sqrt{(-9)^2 + (-6)^2}\)
A1 for awrt 10.8 (exact value \(3\sqrt{13}\)) (N)
Worked solution: \(|\mathbf{F}_3| = \sqrt{(-9)^2 + (-6)^2} = 3\sqrt{13} = 10.8\) N (3 s.f.)
(c)Answer: \(123.7^\circ\) °
M1 for \(\cos\theta = \frac{-6}{|\mathbf{F}_3|}\) or \(\tan^{-1}\frac{9}{6}\) with correct use of the quadrant
A1 for \(123.7^\circ\)
Worked solution: \(\cos\theta = \frac{-6}{3\sqrt{13}} \Rightarrow \theta = 123.7^\circ\) (obtuse, since \(\mathbf{F}_3\) has a negative \(\mathbf{j}\) component)
Question 3Hard8 marks
Relative to a fixed origin \(O\), the points \(A\), \(B\) and \(C\) have coordinates \((1, 2, 3)\), \((4, 0, 7)\) and \((10, -4, 15)\) respectively.
(a) Prove that \(A\), \(B\) and \(C\) are collinear, and find the ratio \(AB : BC\).[3]
(b) The point \(P\) lies on the line through \(A\) and \(B\), with \(\overrightarrow{OP} = \overrightarrow{OA} + \lambda\overrightarrow{AB}\). Show that \(|\overrightarrow{OP}|^2 = 29\lambda^2 + 22\lambda + 14\).[2]
(c) Hence find the exact shortest distance from \(O\) to the line through \(A\) and \(B\).[3]
Show the answer and mark scheme
(a)Answer: \(\overrightarrow{AB} = 3\mathbf{i} - 2\mathbf{j} + 4\mathbf{k}\), \(\overrightarrow{BC} = 6\mathbf{i} - 4\mathbf{j} + 8\mathbf{k} = 2\overrightarrow{AB}\); parallel with the common point \(B\). \(AB : BC = 1 : 2\).
M1 for finding \(\overrightarrow{AB}\) and \(\overrightarrow{BC}\) (or \(\overrightarrow{AC}\))
M1 for \(\overrightarrow{BC} = 2\overrightarrow{AB}\), so they are parallel, and they share the point \(B\)
A1 for collinear, with \(AB : BC = 1 : 2\)
Worked solution: \(\overrightarrow{AB} = \begin{pmatrix} 3 \\ -2 \\ 4 \end{pmatrix}\) and \(\overrightarrow{BC} = \begin{pmatrix} 6 \\ -4 \\ 8 \end{pmatrix} = 2\overrightarrow{AB}\). So \(AB\) and \(BC\) are parallel, and they have the point \(B\) in common, so \(A\), \(B\) and \(C\) lie on one straight line. \(AB : BC = 1 : 2\).
(b)Answer: \(\overrightarrow{OP} = (1 + 3\lambda)\mathbf{i} + (2 - 2\lambda)\mathbf{j} + (3 + 4\lambda)\mathbf{k}\); the sum of squares is \(29\lambda^2 + 22\lambda + 14\).
M1 for \(\overrightarrow{OP} = (1 + 3\lambda, 2 - 2\lambda, 3 + 4\lambda)\) and \(|\overrightarrow{OP}|^2\) as the sum of the squares of the components
A1* for expanding to \(29\lambda^2 + 22\lambda + 14\)