WAEC 2010 · Paper 2 · Q7

Vectors

Vectors $\mathbf p$ , $\mathbf q$ and $\mathbf r$ are given by $\mathbf p = \begin{pmatrix} 2 \\ -3 \end{pmatrix}$ , $\mathbf q = \begin{pmatrix} 4 \\ 2 \end{pmatrix}$ and $\mathbf r = \begin{pmatrix} 3 \\ -2 \end{pmatrix}$ . Find:

(a)
$4\mathbf p - 2\mathbf q + 5\mathbf r$ ;
Separate values with commas, e.g. 3, −2
(b)
the position vector which divides $\mathbf p$ and $\mathbf q$ in the ratio $2 : 3$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

Multiply each vector by its number: $4\mathbf p = \begin{pmatrix} 8 \\ -12 \end{pmatrix}$ , $2\mathbf q = \begin{pmatrix} 8 \\ 4 \end{pmatrix}$ , $5\mathbf r = \begin{pmatrix} 15 \\ -10 \end{pmatrix}$ .
Combine the components: $\begin{pmatrix} 8 - 8 + 15 \\ -12 - 4 - 10 \end{pmatrix} = \begin{pmatrix} 15 \\ -26 \end{pmatrix}$ .

(b)

The point dividing $\mathbf p$ to $\mathbf q$ in the ratio $2 : 3$ has position vector $\dfrac{3\mathbf p + 2\mathbf q}{2 + 3}$ .
Work out the top: $3\begin{pmatrix} 2 \\ -3 \end{pmatrix} + 2\begin{pmatrix} 4 \\ 2 \end{pmatrix} = \begin{pmatrix} 14 \\ -5 \end{pmatrix}$ .
Divide by 5: $\dfrac15\begin{pmatrix} 14 \\ -5 \end{pmatrix} = \begin{pmatrix} 2.8 \\ -1 \end{pmatrix}$ .

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