H2 Math Sun 930am

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 $pi_4$
$\Rightarrow \begin{bmatrix}2.5\\ 1.5\\ 0\end{bmatrix} \bullet \begin{bmatrix}4\\ 5\\ 1\end{bmatrix} = 4k+6$
$k = \frac{23}{8}$

About

KS Teng

KS has been teaching H2/H1 Mathematics and IB mathematics for the more than 10 years. Having taught students from all various Junior Colleges, KS adapts to students' abilities and help them better understand the topics. As someone who loves to teach mathematics and sees it as a truly useful tool in life, KS seeks to enable students to appreciate math. Therefore, his tuition mission is to motivate and cultivate students to be independent and confident thinkers.

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February 23rd, 2016 12:12 AM

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