H2 Math Sun 930am

JC Mathematics

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

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

JC Mathematics

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

MF26


Why is this conclusion valid?
Why is this conclusion valid?

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}

H2 Math Sat 130pm

JC Mathematics

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

MF26


Hence part
Hence part

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


Question 3
Question 3

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

JC Mathematics

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

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

JC Mathematics

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

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}