Vectors (Mathematics Extension 1 content) quiz

$(5, -5)$

$(-6, -4)$

$(1, 3)$

$(1, -3)$

$17$

$\sqrt{17}$

$169$

$13$

$\left(\tfrac{3}{5}, \tfrac{4}{5}\right)$

$(3, 4)$

$(1, 1)$

$\left(\tfrac{1}{3}, \tfrac{1}{4}\right)$

$5$

$(8, -3)$

$11$

$-5$

$|\mathbf{u}| = |\mathbf{v}|$

$\mathbf{u} \cdot \mathbf{v} = 0$

$\mathbf{u} + \mathbf{v} = \mathbf{0}$

$\mathbf{u} = k\mathbf{v}$ for some scalar $k$

$(2, 3)$ and $(-3, 12)$

$(1, 2)$ and $(2, -1)$

$(2, -4)$ and $(-3, 6)$

$(1, 0)$ and $(0, 1)$

$0$

$1$

$\tfrac{\sqrt{3}}{2}$

$\tfrac{1}{\sqrt{2}}$

Acute (strictly between $0^\circ$ and $90^\circ$)

Obtuse (between $90^\circ$ and $180^\circ$)

Right ($90^\circ$)

Zero ($0^\circ$)

$3$

$4$

$5$

$7$

$(0, 3)$

$(4, 3)$

$(4, 0)$

$(1, 0)$

$\mathbf{r}(t) = (3, -1) + t(1, 2)$

$\mathbf{r}(t) = (1, 2) + (3, -1)$

$\mathbf{r}(t) = t(1, 2) + t(3, -1)$

$\mathbf{r}(t) = (1, 2) + t(3, -1)$

$5\mathbf{i} - 6\mathbf{j}$

$5\mathbf{i} + 2\mathbf{j}$

$7\mathbf{i} - 6\mathbf{j}$

$6\mathbf{i} + 2\mathbf{j}$

$|\mathbf{a}| = |\mathbf{b}| = |\mathbf{c}| = |\mathbf{d}|$

$\mathbf{a} \cdot \mathbf{c} = 0$

$\mathbf{a} + \mathbf{c} = \mathbf{b} + \mathbf{d}$

$\mathbf{a} = \mathbf{c}$ and $\mathbf{b} = \mathbf{d}$

$5$

$25$

$\sqrt{25} = 5$ or $-5$

$625$

Write down the final answer straight away with no working at all, since the marker only ever checks the boxed result at the very end and never awards any marks for the method itself.

Draw the two vectors carefully to scale, then read off and guess the projection directly from the measured diagram.

Find the magnitude of each vector separately and add the two magnitudes together to obtain the scalar projection.

Compute $\mathbf{u} \cdot \mathbf{v}$ and $|\mathbf{v}|^2$, then apply $\frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{v}|^2}\mathbf{v}$, stating the formula used.