Vectors: CJC 2024 Promo Q12

(a)

$$\overrightarrow{OA}=\begin{pmatrix}1\\2\\3\end{pmatrix},\quad \overrightarrow{OB}=\begin{pmatrix}2\\-1\\4\end{pmatrix},\quad \overrightarrow{OC}=\begin{pmatrix}4\\5\\-1\end{pmatrix},\quad \overrightarrow{OD}=\begin{pmatrix}-2\\0\\-5\end{pmatrix}$$ $$\begin{aligned} \overrightarrow{AB} &= \begin{pmatrix}2\\-1\\4\end{pmatrix}-\begin{pmatrix}1\\2\\3\end{pmatrix}=\begin{pmatrix}1\\-3\\1\end{pmatrix},\\ \overrightarrow{AC} &= \begin{pmatrix}4\\5\\-1\end{pmatrix}-\begin{pmatrix}1\\2\\3\end{pmatrix}=\begin{pmatrix}3\\3\\-4\end{pmatrix},\\ \overrightarrow{AB} \times \overrightarrow{AC} &= \begin{pmatrix}1\\-3\\1\end{pmatrix}\times\begin{pmatrix}3\\3\\-4\end{pmatrix}\\ &=\begin{pmatrix}(-3)(-4)-(1)(3)\\-[(1)(-4)-(1)(3)]\\(1)(3)-(-3)(3)\end{pmatrix} = \begin{pmatrix}9\\7\\12\end{pmatrix}. \end{aligned}$$ $$\begin{aligned} \Pi:\quad\mathbf{r} \cdot \begin{pmatrix}9\\7\\12\end{pmatrix} &= \begin{pmatrix}1\\2\\3\end{pmatrix} \cdot \begin{pmatrix}9\\7\\12\end{pmatrix}\\ &= 9 + 14 + 36\\ &=59. \end{aligned}$$ $\therefore\ \alpha = 59$

(b)

$$\begin{aligned} \overrightarrow{BD} &= \begin{pmatrix} -2 \\ 0 \\ -5 \end{pmatrix} - \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix} = \begin{pmatrix} -4 \\ 1 \\ -9 \end{pmatrix}. \end{aligned}$$ Method 1⃝: Let $\theta$ be the angle between the line $BD$ and $\Pi$. $$\begin{aligned} \left|\begin{pmatrix}-4\\1\\-9\end{pmatrix}\cdot\begin{pmatrix}9\\7\\12\end{pmatrix}\right|&=\left|\begin{pmatrix}-4\\1\\-9\end{pmatrix}\right|\left|\begin{pmatrix}9\\7\\12\end{pmatrix}\right|\sin\theta\\ 137&=\sqrt{98}\sqrt{274}\sin\theta\\ \sin\theta&=\frac{137}{\sqrt{98}\sqrt{274}}\\ \theta&=56.7^\circ\ (\text{to 1 d.p.})\text{ or }0.990\ (\text{to 3 s.f.}). \end{aligned}$$

Method 2⃝:

Let $\theta$ be the angle between the line $BD$ and normal of $\Pi$. $$\begin{aligned} \left|\begin{pmatrix}-4\\1\\-9\end{pmatrix}\cdot\begin{pmatrix}9\\7\\12\end{pmatrix}\right|&=\left|\begin{pmatrix}-4\\1\\-9\end{pmatrix}\right|\left|\begin{pmatrix}9\\7\\12\end{pmatrix}\right|\cos\theta\\ 137&=\sqrt{98}\sqrt{274}\cos\theta\\ \cos\theta&=-\frac{137}{\sqrt{98}\sqrt{274}}\\ \theta&=33.275^\circ\text{ or }0.58075 \end{aligned}$$ $\therefore$ angle between the line $BD$ and $\Pi$ is $90^\circ-33.275^\circ=56.7^\circ$ or $\dfrac\pi2-0.58075=0.990$ (to 3 s.f.).

(c)

Let $F$ be the foot of perpendicular from the point $D$ to $\Pi$. $$\begin{aligned} l_{DF}:\ \mathbf r &= \begin{pmatrix} -2 \\ 0 \\ -5 \end{pmatrix} + \lambda\begin{pmatrix} 9 \\ 7 \\ 12 \end{pmatrix}, \qquad \lambda \in \mathbb{R}. \end{aligned}$$ Since $F$ lies on line $DF$, $$\begin{aligned} \overrightarrow{OF} &= \begin{pmatrix} -2 + 9\lambda \\ 7\lambda \\ -5 + 12\lambda \end{pmatrix}\quad\text{for some }\lambda. \end{aligned}$$

$$\begin{aligned} \Pi:\quad\mathbf r\cdot\begin{pmatrix}9\\7\\12\end{pmatrix}&=59\\ \begin{pmatrix}-2+9\lambda\\7\lambda\\-5+12\lambda\end{pmatrix}\cdot\begin{pmatrix}9\\7\\12\end{pmatrix}&=59 \end{aligned}$$

$$\begin{aligned} 9(-2 + 9\lambda) + 49\lambda + 12(-5 + 12\lambda) &= 59\\ -18 + 81\lambda + 49\lambda - 60 + 144\lambda &= 59\\ 274\lambda &= 137\\ \lambda &= \frac{1}{2}. \end{aligned}$$ $$\begin{aligned} \overrightarrow{OF} &= \begin{pmatrix} -2 + 9\left(\tfrac{1}{2}\right) \\ 7\left(\tfrac{1}{2}\right) \\ -5 + 12\left(\tfrac{1}{2}\right) \end{pmatrix} = \begin{pmatrix} \tfrac{5}{2} \\ \tfrac{7}{2} \\ 1 \end{pmatrix}. \end{aligned}$$

Alternative Method $$\begin{aligned} \overrightarrow{BD}&=\begin{pmatrix}-4\\1\\-9\end{pmatrix} \end{aligned}$$

$$\begin{aligned} \overrightarrow{FD}&=(\overrightarrow{BD}\cdot\hat{\mathbf n})\hat{\mathbf n}\\ &=\left[\begin{pmatrix}-4\\1\\-9\end{pmatrix}\cdot \frac{\begin{pmatrix}9\\7\\12\end{pmatrix}}{\left|\begin{pmatrix}9\\7\\12\end{pmatrix}\right|}\right] \frac{\begin{pmatrix}9\\7\\12\end{pmatrix}}{\left|\begin{pmatrix}9\\7\\12\end{pmatrix}\right|}\\ &=\left[\begin{pmatrix}-4\\1\\-9\end{pmatrix}\cdot \frac{\begin{pmatrix}9\\7\\12\end{pmatrix}}{\sqrt{274}}\right] \frac{\begin{pmatrix}9\\7\\12\end{pmatrix}}{\sqrt{274}}\\ &=\frac1{274}\left[\begin{pmatrix}-4\\1\\-9\end{pmatrix}\cdot\begin{pmatrix}9\\7\\12\end{pmatrix}\right]\begin{pmatrix}9\\7\\12\end{pmatrix}\\ &=\frac1{274}(-137)\begin{pmatrix}9\\7\\12\end{pmatrix}\\ &=-\frac12\begin{pmatrix}9\\7\\12\end{pmatrix}. \end{aligned}$$ $$\begin{aligned} \overrightarrow{FD}&=\overrightarrow{OD}-\overrightarrow{OF}\\ -\frac12\begin{pmatrix}9\\7\\12\end{pmatrix}&=\begin{pmatrix}-2\\0\\-5\end{pmatrix}-\overrightarrow{OF}\\ \overrightarrow{OF}&=\begin{pmatrix}-2\\0\\-5\end{pmatrix}+\frac12\begin{pmatrix}9\\7\\12\end{pmatrix}\\ &=\begin{pmatrix}5/2\\7/2\\1\end{pmatrix}. \end{aligned}$$

(d)

$$\begin{aligned} \text{Area of triangle } ABC &= \frac{1}{2}\left|\overrightarrow{AB} \times \overrightarrow{AC}\right|\\ &=\frac12\left|\begin{pmatrix}9\\7\\12\end{pmatrix}\right|\\ &=\frac12\sqrt{9^2+7^2+12^2}\\ &=\frac12\sqrt{274}\text{ units}^2. \end{aligned}$$ $$\begin{aligned} \overrightarrow{DF} &= \overrightarrow{OF} - \overrightarrow{OD} = \begin{pmatrix} \tfrac{9}{2} \\ \tfrac{7}{2} \\ 6 \end{pmatrix},\\ \left|\overrightarrow{DF}\right| &= \sqrt{\left(\tfrac{9}{2}\right)^2 + \left(\tfrac{7}{2}\right)^2 + 6^2} = \sqrt{\frac{137}{2}}. \end{aligned}$$ $$\begin{aligned} \text{Volume of pyramid} &= \frac{1}{3}\left(\frac{1}{2}\sqrt{274}\right)\left(\sqrt{\frac{137}{2}}\right) = \frac{137}{6} \text{ units}^3. \end{aligned}$$