Vectors: ACJC 2025 Promo Q10

(a)

With normals $\mathbf{n}_1 = \begin{pmatrix}1\\-1\\1\end{pmatrix}$ and $\mathbf{n}_2 = \begin{pmatrix}1\\1\\1\end{pmatrix}$: $$\begin{aligned} \cos\theta &= \frac{|\mathbf{n}_1\cdot\mathbf{n}_2|}{|\mathbf{n}_1||\mathbf{n}_2|} = \frac{|1 - 1 + 1|}{\sqrt{3}\sqrt{3}} = \frac{1}{3}\\ \theta &= 70.5^\circ. \end{aligned}$$

(b)

The direction of the line of intersection is $\mathbf{n}_1\times\mathbf{n}_2 = \begin{pmatrix}-2\\0\\2\end{pmatrix}$, and $(1,1,0)$ lies on both planes. Hence $$\begin{aligned} \mathbf{r} &= \begin{pmatrix}1\\1\\0\end{pmatrix} + \lambda\begin{pmatrix}-1\\0\\1\end{pmatrix}, \qquad \lambda\in\mathbb{R}. \end{aligned}$$

(c)

The given line $x - 3 = -z + 2,\ y = 1$ has direction $\begin{pmatrix}1\\0\\-1\end{pmatrix}$ and passes through $(3,1,2)$; it is parallel to $l$. With $\overrightarrow{AC} = \begin{pmatrix}5\\0\\3\end{pmatrix}$ joining points on the two lines: $$\begin{aligned} \text{distance} &= \frac{\left|\begin{pmatrix}5\\0\\3\end{pmatrix} \times \begin{pmatrix}1\\0\\-1\end{pmatrix}\right|}{\left|\begin{pmatrix}1\\0\\-1\end{pmatrix}\right|} = \frac{\left|\begin{pmatrix}0\\8\\0\end{pmatrix}\right|}{\sqrt{2}} = \frac{8}{\sqrt{2}} = 4\sqrt{2}. \end{aligned}$$

(d)

Let $\overrightarrow{OC} = \begin{pmatrix}-2+\mu\\1\\-1-\mu\end{pmatrix}$. Since angle $ABC = 90^\circ$, $\overrightarrow{BA}\cdot\overrightarrow{BC} = 0$ with $\overrightarrow{BA} = \begin{pmatrix}-3\\0\\-1\end{pmatrix}$ and $\overrightarrow{BC} = \begin{pmatrix}-3+\mu\\0\\-1-\mu\end{pmatrix}$: $$\begin{aligned} (-3)(-3+\mu) + (-1)(-1-\mu) &= 0\\ 10 - 2\mu &= 0\\ \mu &= 5. \end{aligned}$$ Hence $\overrightarrow{OC} = 3\mathbf{i} + \mathbf{j} - 6\mathbf{k}$.

(e)

The foot $N$ lies on the line $\mathbf{r} = \begin{pmatrix}-2\\1\\-1\end{pmatrix} + \alpha\begin{pmatrix}1\\1\\1\end{pmatrix}$. Substituting into $x + y + z = 2$: $$\begin{aligned} (-2+\alpha) + (1+\alpha) + (-1+\alpha) &= 2\\ \alpha &= \frac{4}{3}. \end{aligned}$$ Therefore $\overrightarrow{ON} = -\dfrac{2}{3}\mathbf{i} + \dfrac{7}{3}\mathbf{j} + \dfrac{1}{3}\mathbf{k}$.

(f)

The plane $p_3$ passes through the midpoint $M$ of $A$ and $N$, with the same normal as $p_2$: $$\begin{aligned} \overrightarrow{OM} &= \frac{1}{2}\left(\overrightarrow{OA} + \overrightarrow{ON}\right) = \begin{pmatrix}-\tfrac{4}{3}\\[2pt]\tfrac{5}{3}\\[2pt]-\tfrac{1}{3}\end{pmatrix}. \end{aligned}$$ Then $\mathbf{r}\cdot\begin{pmatrix}1\\1\\1\end{pmatrix} = -\tfrac{4}{3} + \tfrac{5}{3} - \tfrac{1}{3} = 0$, so $$\begin{aligned} x + y + z &= 0. \end{aligned}$$