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Vectors in problem solvingAQA A-Level Maths: Subtopic test

10 questions, 27 marks

AQA A-Level Maths

Vectors in problem solving

Total 27 marks

Name

Class

Date

  1. 1
    A particle of mass 22 kg is acted on by two forces, F1=(3i+2j)\mathbf F_1=(3\mathbf i+2\mathbf j) N and F2=(−5i+4j)\mathbf F_2=(-5\mathbf i+4\mathbf j) N, where i\mathbf i and j\mathbf j are perpendicular unit vectors.
    (a)
    Find the resultant of F1\mathbf F_1 and F2\mathbf F_2.
    [1 mark]
    • A(−2i+6j)(-2\mathbf i+6\mathbf j) N
    • B(8i−2j)(8\mathbf i-2\mathbf j) N
    • C(−2i−2j)(-2\mathbf i-2\mathbf j) N
    • D(2i−6j)(2\mathbf i-6\mathbf j) N
    (b)
    A third force F3\mathbf F_3 is now applied so that the particle is in equilibrium. Find F3\mathbf F_3.
    [1 mark]
    • A(−2i+6j)(-2\mathbf i+6\mathbf j) N
    • B(2i+6j)(2\mathbf i+6\mathbf j) N
    • C(2i−6j)(2\mathbf i-6\mathbf j) N
    • D0\mathbf 0 N
    (c)
    The force F3\mathbf F_3 is removed, so that only F1\mathbf F_1 and F2\mathbf F_2 act. Find the acceleration of the particle.
    [2 marks]

    Total for question 1: 4 marks

  2. 2
    The points PP, QQ and RR have position vectors p=i+2j\mathbf p=\mathbf i+2\mathbf j, q=3i+8j\mathbf q=3\mathbf i+8\mathbf j and r=ki+14j\mathbf r=k\mathbf i+14\mathbf j, where kk is a constant.
    (a)
    Find PQ→\overrightarrow{PQ}.
    [1 mark]
    • A−2i−6j-2\mathbf i-6\mathbf j
    • B2i+6j2\mathbf i+6\mathbf j
    • C4i+10j4\mathbf i+10\mathbf j
    • D2i+10j2\mathbf i+10\mathbf j
    (b)
    Which statement must be true if PP, QQ and RR lie on a straight line?
    [1 mark]
    • A∣PQ→∣=∣QR→∣|\overrightarrow{PQ}|=|\overrightarrow{QR}|
    • BPQ→\overrightarrow{PQ} is perpendicular to PR→\overrightarrow{PR}
    • CPR→=PQ→\overrightarrow{PR}=\overrightarrow{PQ}
    • DPR→=λPQ→\overrightarrow{PR}=\lambda\overrightarrow{PQ} for some scalar λ\lambda
    (c)
    Given that PP, QQ and RR are collinear, find the value of kk.
    [2 marks]

    Total for question 2: 4 marks

  3. 3
    Relative to a port OO, the unit vectors i\mathbf i and j\mathbf j point due east and due north, and distances are in kilometres. At noon a ship is at the point with position vector (8i+6j)(8\mathbf i+6\mathbf j) km. It moves with constant velocity (−2i+j)(-2\mathbf i+\mathbf j) km h⁻¹.
    (a)
    Find the position vector of the ship at 15:00 and its distance from OO at that time.
    [3 marks]
    (b)
    At time TT hours after noon the ship is due north of the port. Find TT and the distance of the ship from the port at that time.
    [4 marks]

    Total for question 3: 7 marks

  4. 4
    In triangle OABOAB, OA→=a\overrightarrow{OA}=\mathbf a and OB→=b\overrightarrow{OB}=\mathbf b, where a\mathbf a and b\mathbf b are not parallel. The point MM lies on ABAB so that AM:MB=1:3AM:MB=1:3, and CC is the midpoint of OAOA.
    (a)
    (i) Find AB→\overrightarrow{AB}. (ii) Show that OM→=34a+14b\overrightarrow{OM}=\frac34\mathbf a+\frac14\mathbf b. (iii) The point NN is such that ON→=3a+b\overrightarrow{ON}=3\mathbf a+\mathbf b. Show that OO, MM and NN lie on a straight line.
    [6 marks]
    (b)
    The lines OMOM and CBCB meet at the point XX. Find OX→\overrightarrow{OX} in terms of a\mathbf a and b\mathbf b.
    [6 marks]

    Total for question 4: 12 marks

End of questions

Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).