Given the column vectors, p = (-3, 4), q = (16, -4), r = (6, 9) and that a = 2p − 3/4q + 2/3r. Express a as a column vector and hence calculate its magnitude.

The expression for $\mathbf{a}$ is $\mathbf{a} = 2\mathbf{p} - \frac{3}{4}\mathbf{q} + \frac{2}{3}\mathbf{r}$.

$$\mathbf{a} = 2\begin{pmatrix} -3 \\ 4 \end{pmatrix} - \frac{3}{4}\begin{pmatrix} 16 \\ -4 \end{pmatrix} + \frac{2}{3}\begin{pmatrix} 9 \\ 6 \end{pmatrix}$$

$$2\begin{pmatrix} -3 \\ 4 \end{pmatrix} = \begin{pmatrix} 2 \times -3 \\ 2 \times 4 \end{pmatrix} = \begin{pmatrix} -6 \\ 8 \end{pmatrix}$$ $$\frac{3}{4}\begin{pmatrix} 16 \\ -4 \end{pmatrix} = \begin{pmatrix} \frac{3}{4} \times 16 \\ \frac{3}{4} \times -4 \end{pmatrix} = \begin{pmatrix} 12 \\ -3 \end{pmatrix}$$ $$\frac{2}{3}\begin{pmatrix} 9 \\ 6 \end{pmatrix} = \begin{pmatrix} \frac{2}{3} \times 9 \\ \frac{2}{3} \times 6 \end{pmatrix} = \begin{pmatrix} 6 \\ 4 \end{pmatrix}$$

$$\mathbf{a} = \begin{pmatrix} -6 \\ 8 \end{pmatrix} - \begin{pmatrix} 12 \\ -3 \end{pmatrix} + \begin{pmatrix} 6 \\ 4 \end{pmatrix}$$ $$\mathbf{a} = \begin{pmatrix} -6 - 12 + 6 \\ 8 - (-3) + 4 \end{pmatrix}$$ $$\mathbf{a} = \begin{pmatrix} -12 \\ 15 \end{pmatrix}$$

The magnitude of a vector $\begin{pmatrix} x \\ y \end{pmatrix}$ is given by $\sqrt{x^2 + y^2}$.

$$|\mathbf{a}| = \sqrt{(-12)^2 + (15)^2}$$ $$|\mathbf{a}| = \sqrt{144 + 225}$$ $$|\mathbf{a}| = \sqrt{369}$$

$$|\mathbf{a}| = \sqrt{9 \times 41}$$ $$|\mathbf{a}| = 3\sqrt{41}$$

• The column vector $\mathbf{a}$ is $\boxed{\begin{pmatrix} -12 \\ 15 \end{pmatrix}}$
• Its magnitude is $\boxed{3\sqrt{41}}$