Vectors (Year 12 Maths Advanced, 2024 syllabus) quiz

$(1, 3)$

$(5, -5)$

$(-6, -4)$

$(1, -3)$

$17$

$\sqrt{17}$

$13$

$169$

$(3, 4)$

$(1, 1)$

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

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

$5$

$(8, -3)$

$11$

$-5$

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

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

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

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

$(2, 3)$ and $(3, 2)$

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

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

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

$0$

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

$1$

$\tfrac{1}{2}$

Acute (less than $90^\circ$)

Right ($90^\circ$)

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

Zero ($0^\circ$)

$3$

$4$

$5$

$7$

$(4, 0)$

$(0, 3)$

$(4, 3)$

$(1, 0)$

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

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

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

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

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

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

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

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

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

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

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

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

$5$

$25$

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

$625$

Write down the answer and move on.

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.

Draw the vectors, then guess the projection from the diagram.

Find the magnitude of each vector and add them.