### H2 Math Sun 930am

MF26

Vectors Q7 [Homework]
(i)
$\vec{OL} = \begin{bmatrix}2\\ 7\\ -1\end{bmatrix}$
$\vec{OM} = \begin{bmatrix}9\\ 0\\ -8\end{bmatrix}$
Using ratio theorem, $\vec{OP} = \frac{2\vec{OM}+5\vec{OL}}{7} = \begin{bmatrix}4\\ 5\\ -3\end{bmatrix}$
Since $\vec{OP}$ is perpendicular to $\begin{bmatrix}4\\ 1\\ q\end{bmatrix}$
$\Rightarrow \begin{bmatrix}4\\ 5\\ -3\end{bmatrix} \bullet \begin{bmatrix}4\\ 1\\ q\end{bmatrix} = 0$
$q = 7$

(ii)
To be a parallelogram, $\vec{OM} = \vec{LN} = \vec{ON} - \vec{OL}$
$\vec{ON} =\begin{bmatrix}11\\ 7\\ -9\end{bmatrix}$
$Area = |\vec{OM} \times \vec{OL}|$
$= |\begin{bmatrix}56\\ -7\\ 63\end{bmatrix}|$
$= \sqrt{7154} = 7 \sqrt{146} units^2$

(iii)
Let $\vec{OQ} = \begin{bmatrix}x\\ y\\ 0\end{bmatrix}$
Since $|\vec{OQ}| = |\vec{OP}|$
$\sqrt{x^2 + y^2} = \sqrt{50}$ — (1)
$\begin{bmatrix}x\\ y\\ 0\end{bmatrix} \bullet \begin{bmatrix}1\\ 0\\ 0\end{bmatrix} = |\begin{bmatrix}x\\ y\\ 0\end{bmatrix} | |\begin{bmatrix}1\\ 0\\ 0\end{bmatrix} | \mathrm{cos} \theta$ — (2)
Solving, $x = \sqrt{50} \mathrm{cos} \theta = 5 \sqrt{2} \mathrm{cos} \theta$
$y = \sqrt{50} \mathrm{sin} \theta = 5 \sqrt{2} \mathrm{sin} \theta$
$\Rightarrow \vec{OQ} = \begin{bmatrix}{5 \sqrt{2} \mathrm{cos} \theta}\\ {5 \sqrt{2} \mathrm{sin} \theta}\\ 0\end{bmatrix}$

Vectors Q8 [Homework]
(i)
$\vec{OA} = \begin{bmatrix}-5\\ -2\\ 3\end{bmatrix}$
$\vec{OC} = \begin{bmatrix}5\\ 2\\ 6\end{bmatrix}$
$\vec{AC} = \vec{OC} - \vec{OA} = \begin{bmatrix}5\\ 2\\ 6\end{bmatrix} - \begin{bmatrix}-5\\ -2\\ 3\end{bmatrix} = \begin{bmatrix}10\\ 4\\ 3\end{bmatrix}$
$l: r = \begin{bmatrix}5\\ 2\\ 6\end{bmatrix} + \lambda \begin{bmatrix}10\\ 4\\ 3\end{bmatrix}, \lambda \in \mathbb{R}$

(ii)
Let R be the top of the vertical pillar,
$l_{QR}: r = \begin{bmatrix}15\\ 6\\ 0\end{bmatrix} + \mu \begin{bmatrix}0\\ 0\\ 1\end{bmatrix}, \mu \in \mathbb{R}$
Since R is collinear with A and C, R is the intersection of line AC and QR.
$\begin{bmatrix}{5 + 10 \mu}\\ {2 + 4 \mu}\\ {6 + 3 \mu}\end{bmatrix} = \begin{bmatrix}15\\ 6\\ {\mu}\end{bmatrix}$
$\Rightarrow \lambda = 1, \mu = 9$
$\vec{OR} = \begin{bmatrix}15\\ 6\\ 9\end{bmatrix}$, and the height is 9m.

(iii)
$\vec{OD} = \begin{bmatrix}-5\\ 2\\ 6\end{bmatrix}$
$\vec{AD} = \vec{OD} - \vec{OA} = \begin{bmatrix}0\\ 4\\ 3\end{bmatrix}$
$\vec{AX} = (\vec{AD} \bullet \frac{\begin{bmatrix}10\\ 4\\ 3\end{bmatrix}}{| \begin{bmatrix}10\\ 4\\ 3\end{bmatrix}|}) \frac{\begin{bmatrix}10\\ 4\\ 3\end{bmatrix}}{| \begin{bmatrix}10\\ 4\\ 3\end{bmatrix}|}$
$= (\begin{bmatrix}0\\ 4\\ 3\end{bmatrix} \bullet \frac{\begin{bmatrix}10\\ 4\\ 3\end{bmatrix}}{\sqrt{125}}) \frac{\begin{bmatrix}10\\ 4\\ 3\end{bmatrix}}{\sqrt{125}}$
$= \frac{25}{125} \begin{bmatrix}10\\ 4\\ 3\end{bmatrix}$
$= \begin{bmatrix}2\\ 0.8\\ 0.6\end{bmatrix}$
$\vec{OX} = \vec{OA} + \vec{AX} = \begin{bmatrix}-3\\ 1.2\\ 3.6\end{bmatrix}$

Vectors Q9 [Homework]
(i)
$\vec{AB} = \begin{bmatrix}-4\\ 5\\ 3\end{bmatrix}$
$\vec{AC} = \begin{bmatrix}1\\ -3\\ 6\end{bmatrix}$
Normal of $\pi_1, ~n_1=\begin{bmatrix}-4\\ 5\\ 3\end{bmatrix} \times \begin{bmatrix}1\\ -3\\ 6\end{bmatrix} = \begin{bmatrix}-21\\ -21\\ -7\end{bmatrix} = -7 \begin{bmatrix}3\\ 3\\ 1\end{bmatrix}$
$\pi_1: r \bullet \begin{bmatrix}3\\ 3\\ 1\end{bmatrix} = \begin{bmatrix}5\\ -1\\ 0\end{bmatrix} \bullet \begin{bmatrix}3\\ 3\\ 1\end{bmatrix} = 12$

(ii)
Let $\theta$ be the acute angle
$\theta - \mathrm{cos}^{-1} |\frac{\begin{bmatrix}3\\ 3\\ 1\end{bmatrix} \bullet \begin{bmatrix}1\\ -1\\ 1\end{bmatrix}}{\sqrt{19}} \sqrt{3}|$
$\theta = 82.4 ^{\circ}$

(iii)
$3x + 3 y + z = 12$ — (1)
$x - y + z = 1$ — (2)

Using GC, $l: r = \begin{bmatrix}2.5\\ 1.5\\ 0\end{bmatrix} + \lambda \begin{bmatrix}-2\\ 1\\ 3\end{bmatrix}, \lambda \in \mathbb{R}$

(iv)
Let $n_3$ be the normal of $\pi_3$
Length of projection $= |\vec{AB} \times n_3|$
$= \frac{1}{\sqrt{26}} |\begin{bmatrix}4\\ -5\\ 3\end{bmatrix} \times \begin{bmatrix}5\\ -1\\ 0\end{bmatrix}| = 15\sqrt{\frac{3}{26}}$

(v)
Required distance $= \frac{1}{\sqrt{3}} + \frac{2}{\sqrt{3}} = \sqrt{3} units$

(vi)
Let normal of $\pi_4 = n_4 = \begin{bmatrix}-2\\ 1\\ 3\end{bmatrix} \times \begin{bmatrix}1\\ -1\\ 1\end{bmatrix} = \begin{bmatrix}4\\ 5\\ 1\end{bmatrix}$
$\pi_4: r \bullet \begin{bmatrix}4\\ 5\\ 1\end{bmatrix} = 4k+6$
If $\pi_1, \pi_2 \mathrm{~and~} \pi_4$ intersect at l,n$\begin{bmatrix}2.5\\ 1.5\\ 0\end{bmatrix}$ lies on $pi_4$
$\Rightarrow \begin{bmatrix}2.5\\ 1.5\\ 0\end{bmatrix} \bullet \begin{bmatrix}4\\ 5\\ 1\end{bmatrix} = 4k+6$
$k = \frac{23}{8}$

### H2 Math Sat 330pm

This page contains all questions and answers asked by students/ discussions from this class. The most recent questions will be at the top.

MF26

Recall the $a \bullet b = |a| |b| \mathrm{cos} \theta$
If the angle is obtuse $\Rightarrow, \theta \in (\frac{\pi}{2}, \pi)$, then $latex \mathrm{cos} \theta < 0$. Thus, $latex a \bullet b < 0$
Vectors Q7 [Homework]
(i)
$\vec{OL} = \begin{bmatrix}2\\ 7\\ -1\end{bmatrix}$
$\vec{OM} = \begin{bmatrix}9\\ 0\\ -8\end{bmatrix}$
Using ratio theorem, $\vec{OP} = \frac{2\vec{OM}+5\vec{OL}}{7} = \begin{bmatrix}4\\ 5\\ -3\end{bmatrix}$
Since $\vec{OP}$ is perpendicular to $\begin{bmatrix}4\\ 1\\ q\end{bmatrix}$
$\Rightarrow \begin{bmatrix}4\\ 5\\ -3\end{bmatrix} \bullet \begin{bmatrix}4\\ 1\\ q\end{bmatrix} = 0$
$q = 7$

(ii)
To be a parallelogram, $\vec{OM} = \vec{LN} = \vec{ON} - \vec{OL}$
$\vec{ON} =\begin{bmatrix}11\\ 7\\ -9\end{bmatrix}$
$Area = |\vec{OM} \times \vec{OL}|$
$= |\begin{bmatrix}56\\ -7\\ 63\end{bmatrix}|$
$= \sqrt{7154} = 7 \sqrt{146} units^2$

(iii)
Let $\vec{OQ} = \begin{bmatrix}x\\ y\\ 0\end{bmatrix}$
Since $|\vec{OQ}| = |\vec{OP}|$
$\sqrt{x^2 + y^2} = \sqrt{50}$ — (1)
$\begin{bmatrix}x\\ y\\ 0\end{bmatrix} \bullet \begin{bmatrix}1\\ 0\\ 0\end{bmatrix} = |\begin{bmatrix}x\\ y\\ 0\end{bmatrix} | |\begin{bmatrix}1\\ 0\\ 0\end{bmatrix} | \mathrm{cos} \theta$ — (2)
Solving, $x = \sqrt{50} \mathrm{cos} \theta = 5 \sqrt{2} \mathrm{cos} \theta$
$y = \sqrt{50} \mathrm{sin} \theta = 5 \sqrt{2} \mathrm{sin} \theta$
$\Rightarrow \vec{OQ} = \begin{bmatrix}{5 \sqrt{2} \mathrm{cos} \theta}\\ {5 \sqrt{2} \mathrm{sin} \theta}\\ 0\end{bmatrix}$

Vectors Q9 [Homework]
(i)
$\vec{AB} = \begin{bmatrix}-4\\ 5\\ 3\end{bmatrix}$
$\vec{AC} = \begin{bmatrix}1\\ -3\\ 6\end{bmatrix}$
Normal of $\pi_1, ~n_1=\begin{bmatrix}-4\\ 5\\ 3\end{bmatrix} \times \begin{bmatrix}1\\ -3\\ 6\end{bmatrix} = \begin{bmatrix}-21\\ -21\\ -7\end{bmatrix} = -7 \begin{bmatrix}3\\ 3\\ 1\end{bmatrix}$
$\pi_1: r \bullet \begin{bmatrix}3\\ 3\\ 1\end{bmatrix} = \begin{bmatrix}5\\ -1\\ 0\end{bmatrix} \bullet \begin{bmatrix}3\\ 3\\ 1\end{bmatrix} = 12$

(ii)
Let $\theta$ be the acute angle
$\theta - \mathrm{cos}^{-1} |\frac{\begin{bmatrix}3\\ 3\\ 1\end{bmatrix} \bullet \begin{bmatrix}1\\ -1\\ 1\end{bmatrix}}{\sqrt{19}} \sqrt{3}|$
$\theta = 82.4 ^{\circ}$

(iii)
$3x + 3 y + z = 12$ — (1)
$x - y + z = 1$ — (2)

Using GC, $l: r = \begin{bmatrix}2.5\\ 1.5\\ 0\end{bmatrix} + \lambda \begin{bmatrix}-2\\ 1\\ 3\end{bmatrix}, \lambda \in \mathbb{R}$

(iv)
Let $n_3$ be the normal of $\pi_3$
Length of projection $= |\vec{AB} \times n_3|$
$= \frac{1}{\sqrt{26}} |\begin{bmatrix}4\\ -5\\ 3\end{bmatrix} \times \begin{bmatrix}5\\ -1\\ 0\end{bmatrix}| = 15\sqrt{\frac{3}{26}}$

(v)
Required distance $= \frac{1}{\sqrt{3}} + \frac{2}{\sqrt{3}} = \sqrt{3} units$

(vi)
Let normal of $\pi_4 = n_4 = \begin{bmatrix}-2\\ 1\\ 3\end{bmatrix} \times \begin{bmatrix}1\\ -1\\ 1\end{bmatrix} = \begin{bmatrix}4\\ 5\\ 1\end{bmatrix}$
$\pi_4: r \bullet \begin{bmatrix}4\\ 5\\ 1\end{bmatrix} = 4k+6$
If $\pi_1, \pi_2 \mathrm{~and~} \pi_4$ intersect at l,
$\begin{bmatrix}2.5\\ 1.5\\ 0\end{bmatrix}$ lies on $pi_4$
$\Rightarrow \begin{bmatrix}2.5\\ 1.5\\ 0\end{bmatrix} \bullet \begin{bmatrix}4\\ 5\\ 1\end{bmatrix} = 4k+6$
$k = \frac{23}{8}$

### Class Q&A

The following contains questions that students from each class asked. Feel free to discuss further. It is not a tuition schedule, just a mode of communication.

### H2 Math Sat 130pm

MF26

We have that $(2-3i)^2 = -5-12i$
$(z-i+1)^2 = 5+12i$
$(z-i+1)^2 = -(-5-12i)$
$(z-i+1)^2 = -(2-3i)^2$
$(z-i+1)^2 = i^2 (2-3i)^2$
$(z-i+1)^2 = [i(2-3i)]^2$
$(z-i+1)^2 = (2i+3)^2$
$(z-i+1)^2 = (2i+3)^2$
$(z-i+1) = \pm(2i+3)$
$z = \pm(2i+3) - 1 + i$
$z = 2i+3 - 1 + i$ or $z = -(2i+3) - 1 + i$
$z = 3i+2$ or $z = -i+2$

This is a worksheet from class.
There is a slight typo, we need to show that $x^2 (x+2)^2 - 1 = (x+1)^2(x+1+\sqrt{2})(x+1-\sqrt{2})$
So for all proof/ show questions, it is important that we can identify whether we should prove from LHS to RHS or RHS to LHS.
For this question, we observe that if we start from LHS, we are trying to expand our expression. But if we start from RHS, we are trying to simplify our expression. It will be generally easier to simplify.

$RHS$
$= (x+1)^2(x+1+\sqrt{2})(x+1-\sqrt{2})$
$= (x+1)^2[(x+1)^2 - (\sqrt{2})^2]$
$= (x+1)^2[(x+1)^2 - 2]$
$= (x+1)^2(x^2 + 2x + 1 - 2]$
$= (x^2 + 2x + 1)(x^2 + 2x -1)$; we can also expand it fully
$= [x(x+2)+1][x(x+2)-1]$
$= x^2 (x+2)^2 -1$

Here’s a much faster but less obvious method!
$LHS$
$= x^2 (x+2)^2 -1$
$= [x(x+2)+1][x(x+2)-1]$
$= (x^2 + 2x + 1)(x^2 + 2x -1)$; using quadratic formula
$= (x+1)^2 (x - \frac{-2-\sqrt{8}}{2})x - \frac{-2+\sqrt{8}}{2})$
$= (x+)^2(x+1+1+\sqrt{2})(x+1-\sqrt{2})$

### H2 Math Mon 730pm

MF26

Vectors Q7 [Homework]
(i)
$\vec{OL} = \begin{bmatrix}2\\ 7\\ -1\end{bmatrix}$
$\vec{OM} = \begin{bmatrix}9\\ 0\\ -8\end{bmatrix}$
Using ratio theorem, $\vec{OP} = \frac{2\vec{OM}+5\vec{OL}}{7} = \begin{bmatrix}4\\ 5\\ -3\end{bmatrix}$
Since $\vec{OP}$ is perpendicular to $\begin{bmatrix}4\\ 1\\ q\end{bmatrix}$
$\Rightarrow \begin{bmatrix}4\\ 5\\ -3\end{bmatrix} \bullet \begin{bmatrix}4\\ 1\\ q\end{bmatrix} = 0$
$q = 7$

(ii)
To be a parallelogram, $\vec{OM} = \vec{LN} = \vec{ON} - \vec{OL}$
$\vec{ON} =\begin{bmatrix}11\\ 7\\ -9\end{bmatrix}$
$Area = |\vec{OM} \times \vec{OL}|$
$= |\begin{bmatrix}56\\ -7\\ 63\end{bmatrix}|$
$= \sqrt{7154} = 7 \sqrt{146} units^2$

(iii)
Let $\vec{OQ} = \begin{bmatrix}x\\ y\\ 0\end{bmatrix}$
Since $|\vec{OQ}| = |\vec{OP}|$
$\sqrt{x^2 + y^2} = \sqrt{50}$ — (1)
$\begin{bmatrix}x\\ y\\ 0\end{bmatrix} \bullet \begin{bmatrix}1\\ 0\\ 0\end{bmatrix} = |\begin{bmatrix}x\\ y\\ 0\end{bmatrix} | |\begin{bmatrix}1\\ 0\\ 0\end{bmatrix} | \mathrm{cos} \theta$ — (2)
Solving, $x = \sqrt{50} \mathrm{cos} \theta = 5 \sqrt{2} \mathrm{cos} \theta$
$y = \sqrt{50} \mathrm{sin} \theta = 5 \sqrt{2} \mathrm{sin} \theta$
$\Rightarrow \vec{OQ} = \begin{bmatrix}{5 \sqrt{2} \mathrm{cos} \theta}\\ {5 \sqrt{2} \mathrm{sin} \theta}\\ 0\end{bmatrix}$

Vectors Q8 [Homework]
(i)
$\vec{OA} = \begin{bmatrix}-5\\ -2\\ 3\end{bmatrix}$
$\vec{OC} = \begin{bmatrix}5\\ 2\\ 6\end{bmatrix}$
$\vec{AC} = \vec{OC} - \vec{OA} = \begin{bmatrix}5\\ 2\\ 6\end{bmatrix} - \begin{bmatrix}-5\\ -2\\ 3\end{bmatrix} = \begin{bmatrix}10\\ 4\\ 3\end{bmatrix}$
$l: r = \begin{bmatrix}5\\ 2\\ 6\end{bmatrix} + \lambda \begin{bmatrix}10\\ 4\\ 3\end{bmatrix}, \lambda \in \mathbb{R}$

(ii)
Let R be the top of the vertical pillar,
$l_{QR}: r = \begin{bmatrix}15\\ 6\\ 0\end{bmatrix} + \mu \begin{bmatrix}0\\ 0\\ 1\end{bmatrix}, \mu \in \mathbb{R}$
Since R is collinear with A and C, R is the intersection of line AC and QR.
$\begin{bmatrix}{5 + 10 \mu}\\ {2 + 4 \mu}\\ {6 + 3 \mu}\end{bmatrix} = \begin{bmatrix}15\\ 6\\ {\mu}\end{bmatrix}$
$\Rightarrow \lambda = 1, \mu = 9$
$\vec{OR} = \begin{bmatrix}15\\ 6\\ 9\end{bmatrix}$, and the height is 9m.

(iii)
$\vec{OD} = \begin{bmatrix}-5\\ 2\\ 6\end{bmatrix}$
$\vec{AD} = \vec{OD} - \vec{OA} = \begin{bmatrix}0\\ 4\\ 3\end{bmatrix}$
$\vec{AX} = (\vec{AD} \bullet \frac{\begin{bmatrix}10\\ 4\\ 3\end{bmatrix}}{| \begin{bmatrix}10\\ 4\\ 3\end{bmatrix}|}) \frac{\begin{bmatrix}10\\ 4\\ 3\end{bmatrix}}{| \begin{bmatrix}10\\ 4\\ 3\end{bmatrix}|}$
$= (\begin{bmatrix}0\\ 4\\ 3\end{bmatrix} \bullet \frac{\begin{bmatrix}10\\ 4\\ 3\end{bmatrix}}{\sqrt{125}}) \frac{\begin{bmatrix}10\\ 4\\ 3\end{bmatrix}}{\sqrt{125}}$
$= \frac{25}{125} \begin{bmatrix}10\\ 4\\ 3\end{bmatrix}$
$= \begin{bmatrix}2\\ 0.8\\ 0.6\end{bmatrix}$
$\vec{OX} = \vec{OA} + \vec{AX} = \begin{bmatrix}-3\\ 1.2\\ 3.6\end{bmatrix}$

Vectors Q9 [Homework]
(i)
$\vec{AB} = \begin{bmatrix}-4\\ 5\\ 3\end{bmatrix}$
$\vec{AC} = \begin{bmatrix}1\\ -3\\ 6\end{bmatrix}$
Normal of $\pi_1, ~n_1=\begin{bmatrix}-4\\ 5\\ 3\end{bmatrix} \times \begin{bmatrix}1\\ -3\\ 6\end{bmatrix} = \begin{bmatrix}-21\\ -21\\ -7\end{bmatrix} = -7 \begin{bmatrix}3\\ 3\\ 1\end{bmatrix}$
$\pi_1: r \bullet \begin{bmatrix}3\\ 3\\ 1\end{bmatrix} = \begin{bmatrix}5\\ -1\\ 0\end{bmatrix} \bullet \begin{bmatrix}3\\ 3\\ 1\end{bmatrix} = 12$

(ii)
Let $\theta$ be the acute angle
$\theta - \mathrm{cos}^{-1} |\frac{\begin{bmatrix}3\\ 3\\ 1\end{bmatrix} \bullet \begin{bmatrix}1\\ -1\\ 1\end{bmatrix}}{\sqrt{19}} \sqrt{3}|$
$\theta = 82.4 ^{\circ}$

(iii)
$3x + 3 y + z = 12$ — (1)
$x - y + z = 1$ — (2)

Using GC, $l: r = \begin{bmatrix}2.5\\ 1.5\\ 0\end{bmatrix} + \lambda \begin{bmatrix}-2\\ 1\\ 3\end{bmatrix}, \lambda \in \mathbb{R}$

(iv)
Let $n_3$ be the normal of $\pi_3$
Length of projection $= |\vec{AB} \times n_3|$
$= \frac{1}{\sqrt{26}} |\begin{bmatrix}4\\ -5\\ 3\end{bmatrix} \times \begin{bmatrix}5\\ -1\\ 0\end{bmatrix}| = 15\sqrt{\frac{3}{26}}$

(v)
Required distance $= \frac{1}{\sqrt{3}} + \frac{2}{\sqrt{3}} = \sqrt{3} units$

(vi)
Let normal of $\pi_4 = n_4 = \begin{bmatrix}-2\\ 1\\ 3\end{bmatrix} \times \begin{bmatrix}1\\ -1\\ 1\end{bmatrix} = \begin{bmatrix}4\\ 5\\ 1\end{bmatrix}$
$\pi_4: r \bullet \begin{bmatrix}4\\ 5\\ 1\end{bmatrix} = 4k+6$
If $\pi_1, \pi_2 \mathrm{~and~} \pi_4$ intersect at l,n$\begin{bmatrix}2.5\\ 1.5\\ 0\end{bmatrix}$ lies on $pi_4$
$\Rightarrow \begin{bmatrix}2.5\\ 1.5\\ 0\end{bmatrix} \bullet \begin{bmatrix}4\\ 5\\ 1\end{bmatrix} = 4k+6$
$k = \frac{23}{8}$

### H2 Math Mon 2pm

MF26

Vectors Q7 [Homework]
(i)
$\vec{OL} = \begin{bmatrix}2\\ 7\\ -1\end{bmatrix}$
$\vec{OM} = \begin{bmatrix}9\\ 0\\ -8\end{bmatrix}$
Using ratio theorem, $\vec{OP} = \frac{2\vec{OM}+5\vec{OL}}{7} = \begin{bmatrix}4\\ 5\\ -3\end{bmatrix}$
Since $\vec{OP}$ is perpendicular to $\begin{bmatrix}4\\ 1\\ q\end{bmatrix}$
$\Rightarrow \begin{bmatrix}4\\ 5\\ -3\end{bmatrix} \bullet \begin{bmatrix}4\\ 1\\ q\end{bmatrix} = 0$
$q = 7$

(ii)
To be a parallelogram, $\vec{OM} = \vec{LN} = \vec{ON} - \vec{OL}$
$\vec{ON} =\begin{bmatrix}11\\ 7\\ -9\end{bmatrix}$
$Area = |\vec{OM} \times \vec{OL}|$
$= |\begin{bmatrix}56\\ -7\\ 63\end{bmatrix}|$
$= \sqrt{7154} = 7 \sqrt{146} units^2$

(iii)
Let $\vec{OQ} = \begin{bmatrix}x\\ y\\ 0\end{bmatrix}$
Since $|\vec{OQ}| = |\vec{OP}|$
$\sqrt{x^2 + y^2} = \sqrt{50}$ — (1)
$\begin{bmatrix}x\\ y\\ 0\end{bmatrix} \bullet \begin{bmatrix}1\\ 0\\ 0\end{bmatrix} = |\begin{bmatrix}x\\ y\\ 0\end{bmatrix} | |\begin{bmatrix}1\\ 0\\ 0\end{bmatrix} | \mathrm{cos} \theta$ — (2)
Solving, $x = \sqrt{50} \mathrm{cos} \theta = 5 \sqrt{2} \mathrm{cos} \theta$
$y = \sqrt{50} \mathrm{sin} \theta = 5 \sqrt{2} \mathrm{sin} \theta$
$\Rightarrow \vec{OQ} = \begin{bmatrix}{5 \sqrt{2} \mathrm{cos} \theta}\\ {5 \sqrt{2} \mathrm{sin} \theta}\\ 0\end{bmatrix}$

Vectors Q8 [Homework]
(i)
$\vec{OA} = \begin{bmatrix}-5\\ -2\\ 3\end{bmatrix}$
$\vec{OC} = \begin{bmatrix}5\\ 2\\ 6\end{bmatrix}$
$\vec{AC} = \vec{OC} - \vec{OA} = \begin{bmatrix}5\\ 2\\ 6\end{bmatrix} - \begin{bmatrix}-5\\ -2\\ 3\end{bmatrix} = \begin{bmatrix}10\\ 4\\ 3\end{bmatrix}$
$l: r = \begin{bmatrix}5\\ 2\\ 6\end{bmatrix} + \lambda \begin{bmatrix}10\\ 4\\ 3\end{bmatrix}, \lambda \in \mathbb{R}$

(ii)
Let R be the top of the vertical pillar,
$l_{QR}: r = \begin{bmatrix}15\\ 6\\ 0\end{bmatrix} + \mu \begin{bmatrix}0\\ 0\\ 1\end{bmatrix}, \mu \in \mathbb{R}$
Since R is collinear with A and C, R is the intersection of line AC and QR.
$\begin{bmatrix}{5 + 10 \mu}\\ {2 + 4 \mu}\\ {6 + 3 \mu}\end{bmatrix} = \begin{bmatrix}15\\ 6\\ {\mu}\end{bmatrix}$
$\Rightarrow \lambda = 1, \mu = 9$
$\vec{OR} = \begin{bmatrix}15\\ 6\\ 9\end{bmatrix}$, and the height is 9m.

(iii)
$\vec{OD} = \begin{bmatrix}-5\\ 2\\ 6\end{bmatrix}$
$\vec{AD} = \vec{OD} - \vec{OA} = \begin{bmatrix}0\\ 4\\ 3\end{bmatrix}$
$\vec{AX} = (\vec{AD} \bullet \frac{\begin{bmatrix}10\\ 4\\ 3\end{bmatrix}}{| \begin{bmatrix}10\\ 4\\ 3\end{bmatrix}|}) \frac{\begin{bmatrix}10\\ 4\\ 3\end{bmatrix}}{| \begin{bmatrix}10\\ 4\\ 3\end{bmatrix}|}$
$= (\begin{bmatrix}0\\ 4\\ 3\end{bmatrix} \bullet \frac{\begin{bmatrix}10\\ 4\\ 3\end{bmatrix}}{\sqrt{125}}) \frac{\begin{bmatrix}10\\ 4\\ 3\end{bmatrix}}{\sqrt{125}}$
$= \frac{25}{125} \begin{bmatrix}10\\ 4\\ 3\end{bmatrix}$
$= \begin{bmatrix}2\\ 0.8\\ 0.6\end{bmatrix}$
$\vec{OX} = \vec{OA} + \vec{AX} = \begin{bmatrix}-3\\ 1.2\\ 3.6\end{bmatrix}$

Vectors Q9 [Homework]
(i)
$\vec{AB} = \begin{bmatrix}-4\\ 5\\ 3\end{bmatrix}$
$\vec{AC} = \begin{bmatrix}1\\ -3\\ 6\end{bmatrix}$
Normal of $\pi_1, ~n_1=\begin{bmatrix}-4\\ 5\\ 3\end{bmatrix} \times \begin{bmatrix}1\\ -3\\ 6\end{bmatrix} = \begin{bmatrix}-21\\ -21\\ -7\end{bmatrix} = -7 \begin{bmatrix}3\\ 3\\ 1\end{bmatrix}$
$\pi_1: r \bullet \begin{bmatrix}3\\ 3\\ 1\end{bmatrix} = \begin{bmatrix}5\\ -1\\ 0\end{bmatrix} \bullet \begin{bmatrix}3\\ 3\\ 1\end{bmatrix} = 12$

(ii)
Let $\theta$ be the acute angle
$\theta - \mathrm{cos}^{-1} |\frac{\begin{bmatrix}3\\ 3\\ 1\end{bmatrix} \bullet \begin{bmatrix}1\\ -1\\ 1\end{bmatrix}}{\sqrt{19}} \sqrt{3}|$
$\theta = 82.4 ^{\circ}$

(iii)
$3x + 3 y + z = 12$ — (1)
$x - y + z = 1$ — (2)

Using GC, $l: r = \begin{bmatrix}2.5\\ 1.5\\ 0\end{bmatrix} + \lambda \begin{bmatrix}-2\\ 1\\ 3\end{bmatrix}, \lambda \in \mathbb{R}$

(iv)
Let $n_3$ be the normal of $\pi_3$
Length of projection $= |\vec{AB} \times n_3|$
$= \frac{1}{\sqrt{26}} |\begin{bmatrix}4\\ -5\\ 3\end{bmatrix} \times \begin{bmatrix}5\\ -1\\ 0\end{bmatrix}| = 15\sqrt{\frac{3}{26}}$

(v)
Required distance $= \frac{1}{\sqrt{3}} + \frac{2}{\sqrt{3}} = \sqrt{3} units$

(vi)
Let normal of $\pi_4 = n_4 = \begin{bmatrix}-2\\ 1\\ 3\end{bmatrix} \times \begin{bmatrix}1\\ -1\\ 1\end{bmatrix} = \begin{bmatrix}4\\ 5\\ 1\end{bmatrix}$
$\pi_4: r \bullet \begin{bmatrix}4\\ 5\\ 1\end{bmatrix} = 4k+6$
If $\pi_1, \pi_2 \mathrm{~and~} \pi_4$ intersect at l,n$\begin{bmatrix}2.5\\ 1.5\\ 0\end{bmatrix}$ lies on $pi_4$
$\Rightarrow \begin{bmatrix}2.5\\ 1.5\\ 0\end{bmatrix} \bullet \begin{bmatrix}4\\ 5\\ 1\end{bmatrix} = 4k+6$
$k = \frac{23}{8}$