Stepwise
Start practisingPractise

WAEC 2009 · Paper 2 · Q7

The position vectors of points PP and RR are (4i−3j)(4\mathbf i - 3\mathbf j) and (8i+j)(8\mathbf i + \mathbf j) respectively. Find:

  1. (a)

    the unit vector in the direction of (4i−3j)(4\mathbf i - 3\mathbf j);

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

  2. (b)

    correct to one decimal place, the acute angle between the two vectors.

Worked solution (try it first)

(a)

  1. The magnitude is ∣4i−3j∣=42+32|4\mathbf i - 3\mathbf j| = \sqrt{4^2 + 3^2}, which is 25=5\sqrt{25} = 5.
  2. Divide the vector by its length: 15(4i−3j)=45i−35j\frac15(4\mathbf i - 3\mathbf j) = \frac45\mathbf i - \frac35\mathbf j.

(b)

  1. Dot product: (4i−3j)⋅(8i+j)=32−3(4\mathbf i - 3\mathbf j)\cdot(8\mathbf i + \mathbf j) = 32 - 3
    =29= 29.
  2. ∣8i+j∣=64+1|8\mathbf i + \mathbf j| = \sqrt{64 + 1}
    =65= \sqrt{65}.
  3. cos⁡θ=29565\cos\theta = \dfrac{29}{5\sqrt{65}}
    =0.7194= 0.7194.
  4. So θ=44.0∘\theta = 44.0^\circ (1 d.p.).

Report a problem with this question