Vectors: ACJC 2025 Promo Q7

(a)

Since $\mathbf{a}+\mathbf{b}-\mathbf{c}$ and $\mathbf{a}-\mathbf{b}+\mathbf{c}$ are parallel, $$\begin{aligned} (\mathbf{a}+\mathbf{b}-\mathbf{c}) \times (\mathbf{a}-\mathbf{b}+\mathbf{c}) &= \mathbf{0}\\ -2(\mathbf{a}\times\mathbf{b}) + 2(\mathbf{a}\times\mathbf{c}) &= \mathbf{0}\\ \mathbf{a}\times(\mathbf{c}-\mathbf{b}) &= \mathbf{0}. \end{aligned}$$ Since $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{BC} = \mathbf{c}-\mathbf{b}$, we have $\overrightarrow{OA}\times\overrightarrow{BC} = \mathbf{0}$, so $\overrightarrow{OA}$ and $\overrightarrow{BC}$ are parallel. (shown)

(b)

With $\overrightarrow{BO} = -\mathbf{b}$, $\overrightarrow{BC} = \mathbf{c}-\mathbf{b}$, $|\mathbf{b}| = 1$ and $|\overrightarrow{BC}| = 2|\overrightarrow{OA}| = 2$: $$\begin{aligned} \overrightarrow{BO}\cdot\overrightarrow{BC} &= |\overrightarrow{BO}|\,|\overrightarrow{BC}|\cos 60^\circ\\ (-\mathbf{b})\cdot(\mathbf{c}-\mathbf{b}) &= (1)(2)\left(\tfrac{1}{2}\right)\\ -\mathbf{b}\cdot\mathbf{c} + |\mathbf{b}|^2 &= 1\\ \mathbf{b}\cdot\mathbf{c} &= 0. \end{aligned}$$ Hence $\overrightarrow{OB}$ and $\overrightarrow{OC}$ are perpendicular. (shown)

(c)

Since $\mathbf{h} = \mathbf{b}\times\mathbf{c}$, $\mathbf{h}$ is perpendicular to both $\mathbf{b}$ and $\mathbf{c}$, so $\left|\overrightarrow{OH}\right|$ is the height of the pyramid. As $\mathbf{b}\cdot\mathbf{c} = 0$, the angle between $\mathbf{b}$ and $\mathbf{c}$ is $90^\circ$: $$\begin{aligned} \left|\overrightarrow{OH}\right| &= |\mathbf{b}\times\mathbf{c}| = |\mathbf{b}||\mathbf{c}|\sin 90^\circ = (1)(\sqrt{3})(1) = \sqrt{3}. \end{aligned}$$ Triangle $OBC$ is right-angled at $O$, so $$\begin{aligned} \text{Area of } \triangle OBC &= \frac{1}{2}|\mathbf{b}\times\mathbf{c}| = \frac{\sqrt{3}}{2}. \end{aligned}$$ Therefore $$\begin{aligned} \text{Volume} &= \frac{1}{3}\left(\frac{\sqrt{3}}{2}\right)(\sqrt{3}) = \frac{1}{2} \text{ units}^3. \end{aligned}$$

When finding $|\mathbf{b}\times\mathbf{c}|$, use $\sin 90^\circ$ (since $\mathbf{b}\cdot\mathbf{c}=0$ gives angle $90^\circ$), not $\sin 60^\circ$.