H2 Math Sun 1130am

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}

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