WAEC 2018 · Paper 2 · Q15

  1. (a)

    If a=6i−5j\mathbf a = 6\mathbf i - 5\mathbf j, b=2i+7j\mathbf b = 2\mathbf i + 7\mathbf j and c=8i+2j\mathbf c = 8\mathbf i + 2\mathbf j, find the values of the scalars xx and yy such that c=xa+yb\mathbf c = x\mathbf a + y\mathbf b.

    Separate values with commas, e.g. 3, −2

  2. (b)

    The vectors a\mathbf a and b\mathbf b are such that ∣a∣=3 cm|\mathbf a| = 3\text{ cm}, ∣b∣=10 cm|\mathbf b| = 10\text{ cm} and ∣a+b∣=139|\mathbf a + \mathbf b| = \sqrt{139}. Find: (i) the angle between a\mathbf a and b\mathbf b; (ii) the scalar (dot) product of a\mathbf a and b\mathbf b.

    Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

  1. Match the parts: 6x+2y=86x + 2y = 8 and −5x+7y=2-5x + 7y = 2.
  2. From the first, y=4−3xy = 4 - 3x.
  3. Substitute: −5x+28−21x=2-5x + 28 - 21x = 2, so x=1x = 1 and y=1y = 1.

(b)(i)

  1. ∣a+b∣2=∣a∣2+∣b∣2+2∣a∣∣b∣cos⁡θ|\mathbf a + \mathbf b|^2 = |\mathbf a|^2 + |\mathbf b|^2 + 2|\mathbf a||\mathbf b|\cos\theta.
  2. 139=9+100+60cos⁡θ139 = 9 + 100 + 60\cos\theta, so cos⁡θ=3060=12\cos\theta = \frac{30}{60} = \frac12 and θ=60∘\theta = 60^\circ.

(ii)

  1. a⋅b=∣a∣∣b∣cos⁡θ\mathbf a \cdot \mathbf b = |\mathbf a||\mathbf b|\cos\theta
    =3×10×12= 3 \times 10 \times \frac12
    =15= 15.

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