Vectors: NYJC 2024 Promo Q5

(a)

$$\begin{aligned} \mathbf a\times\mathbf b&=\begin{pmatrix}3\\2\\-1\end{pmatrix}\times \begin{pmatrix}4\\-3\\2\end{pmatrix}=\begin{pmatrix}1\\-10\\-17\end{pmatrix} \end{aligned}$$ Area of $\triangle AOB$ $$\begin{aligned} &=\frac12\lvert\mathbf a\times\mathbf b\rvert\\ &=\frac12\left\lvert\begin{pmatrix}1\\-10\\-17\end{pmatrix}\right\rvert\\ &=\frac12\sqrt{1+100+289}\\ &=\frac12\sqrt{390}\text{ units}^2 \end{aligned}$$

(b)

Method 1:

Perpendicular ht of tetrahedron = $$\begin{aligned} \frac{1}{\sqrt{1+100+289}}\overrightarrow{OC}\mathbin{\boldsymbol\cdot} \begin{pmatrix}1\\-10\\-17\end{pmatrix} &=\frac{1}{\sqrt{390}}\begin{pmatrix}3\\-1\\-1\end{pmatrix}\mathbin{\boldsymbol\cdot} \begin{pmatrix}1\\-10\\-17\end{pmatrix}\\ &=\frac{30}{\sqrt{390}} \end{aligned}$$ $$\text{Vol of tetrahedron }OABC=\frac13\left(\frac{\sqrt{390}}2\right) \left(\frac{30}{\sqrt{390}}\right)=5\text{ units}^3$$

Method 2

Let $X$ be the foot of perpendicular from $C$ to plane $OAB$. $$\overrightarrow{OX}=\begin{pmatrix}3\\-1\\-1\end{pmatrix} +\mu\begin{pmatrix}1\\-10\\-17\end{pmatrix},\text{ for some }\mu\in\mathbb R$$ Equation of plane $OAB$: $$\mathbf r\mathbin{\boldsymbol\cdot}\begin{pmatrix}1\\-10\\-17\end{pmatrix}=0$$ Since the point $X$ also lies on the plane $OAB$, $$\begin{aligned} \left(\begin{pmatrix}3\\-1\\-1\end{pmatrix}+\mu\begin{pmatrix}1\\-10\\-17\end{pmatrix}\right) \mathbin{\boldsymbol\cdot}\begin{pmatrix}1\\-10\\-17\end{pmatrix}&=0\\ 3+\mu+10+100\mu+17+289\mu&=0\\ 390\mu&=-30\\ \mu&=-\frac1{13} \end{aligned}$$ $$\begin{aligned} \overrightarrow{CX}&=\overrightarrow{OX}-\overrightarrow{OC}\\ &=\begin{pmatrix}3\\-1\\-1\end{pmatrix}-\frac1{13}\begin{pmatrix}1\\-10\\-17\end{pmatrix} -\begin{pmatrix}3\\-1\\-1\end{pmatrix}\\ &=-\frac1{13}\begin{pmatrix}1\\-10\\-17\end{pmatrix} \end{aligned}$$ $$\begin{aligned} \text{perpendicular height, }CX&=\frac1{13}\sqrt{1+(-10)^2+(-17)^2}\\ &=\frac{\sqrt{390}}{13}\\&=\frac{30}{\sqrt{390}} \end{aligned}$$

(c)

Method 1

Perpendicular distance from $B$ to line $OA$ $$\begin{aligned} &=\lvert\mathbf b\times\hat{\mathbf a}\rvert\\ &=\frac{\sqrt{390}}{\sqrt{9+4+1}}\\ &=\sqrt{\frac{195}{7}}\quad\text{or }5.28 \end{aligned}$$

Method 2

$$\text{Area of }OAB=\frac12\sqrt{390}=\frac12OA(h)$$ Thus, perpendicular distance from $B$ to line $OA$ $$h=\frac{\sqrt{390}}{\sqrt{14}}=\sqrt{\frac{195}{7}}.$$