DSE Math M2 2019 ANS

Posted: April 23, 2020 in Mathematics
Tags: DSE, exams, math, solution

DSE_M2_2019_Q1

$\displaystyle f(1+h)-f(1) = \frac{10(1+h)}{7+3(1+h)^2}-\frac{10}{7+3} = \frac{10+10h}{10+6h+3h^2}-\frac{10+6h+3h^2}{10+6h+3h^2}$

$\displaystyle =\frac{4h-3h^2}{10+6h+3h^2}$

$\displaystyle f'(1) = \lim_{h\to0} \frac{f(1+h)-f(1)}{h} = \lim_{h\to0} \frac{1}{h}\cdot\frac{4h-3h^2}{10+6h+3h^2} = \lim_{h\to0} \frac{4-3h}{10+6h+3h^2}$

$\displaystyle =\frac{4}{10} = \frac25$

DSE_M2_2019_Q2

(a)

$P(x) = (x+\lambda)((x+\lambda)^5-15)+4(3-2(x+\lambda)^2)$

$= (x+\lambda)^6-15(x+\lambda)+12-8(x+\lambda)^2$

$x^3$ term: $\displaystyle \binom{6}{3} x^{6-3} \lambda^3 = \frac{6\cdot5\cdot4}{3\cdot2}\lambda^3x^3 = 160x^3$

$20\lambda^3=160$

$\lambda^3=8$

$\lambda = 2$

(b)

$P'(x) = 6(x+2)^5-15-16(x+2)$

$P'(0) = 6(2)^5 - 15 - 16(2) = 6\cdot32-15-32 = 5\cdot32-15 = 160-15=145$

DSE_M2_2019_Q3

(a)

$\displaystyle V=-2\int t \text{ d}t = -t^2 + C$

$\displaystyle V(0) = C = 580 \quad\Rightarrow \quad V=-t^2+580$

$V(24) = -24^2+580 = 4$

The claim is correct because the vessel has 4 cm³ of liquid X at the end of the experiment.

(b)

$V=h^2+24h, \quad V'=2hh'+24h'$

$\displaystyle h'=\frac{V'}{2h+24}$

When t = 18,

$V(18) = h^2 + 24h$

$-18^2+580 = h^2 + 24h$

$h^2+24h-256=0$

$(h+32)(h-8)=0$

$h=8 \quad (\text{reject } h=-32)$

Then,

$\displaystyle \frac{\text{d}h}{\text{d}t}\bigg|_{t=18} = \frac{-2(18)}{2(8)+24} = -\frac{9}{10}$

DSE_M2_2019_Q4

(a)

$\displaystyle g'(x) = \frac{\frac1x \cdot \sqrt{x} - \ln x \cdot\frac{1}{2\sqrt{x}}}{x} = \frac{1}{x}\cdot\left(\frac{1}{\sqrt{x}}-\frac{\ln x}{2\sqrt{x}}\right) = \frac{2-\ln x}{2x\sqrt{x}}$

Let $g'(x)=0$

$2-\ln x = 0$

$x=e^2$

If $0<x<e^2 \Rightarrow \ln x < 2 \Rightarrow 2-\ln x > 0 \Rightarrow g'(x)>0$

If $99 > x > e^2 \Rightarrow \ln x > 2 \Rightarrow 2-\ln x < 0 \Rightarrow g'(x)<0$

Therefore, G has only one maximum point.

(b)

$\displaystyle \text{Volume} = \pi\int_1^{e^2} \left(\frac{\ln x}{\sqrt{x}}\right)^2 \text{ d}x = \pi\int_1^{e^2} \frac{\ln^2 x}{x} \text{ d}x$

Let $w=\ln x, \quad \text{d}w=\dfrac{1}{x}\text{ d}x$

$\displaystyle = \pi \int_0^2 w^2 \text{ d}w = \frac{\pi}{3} w^3 \bigg|_0^2 = \frac{8\pi}{3}$

DSE_M2_2019_Q5

(a)

Let $\displaystyle P(n): \quad \sum_{k=n}^{2n} \frac{1}{k(k+1)} = \frac{n+1}{n(2n+1)}$

P(1):

$\displaystyle LS = \sum_{k=1}^2 \frac{1}{k(k+1)} = \frac12 + \frac{1}{2(3)} = \frac46 = \frac23$

$\displaystyle RS = \frac{1+1}{1(3)} = \frac23$

Hence, P(1) is true.

Assume P(m) is true, then P(m+1):

$\displaystyle \sum_{k=m+1}^{2(m+1)} \frac{1}{k(k+1)} = \sum_{k=m}^{2m}\frac{1}{k(k+1)} - \frac{1}{m(m+1)} + \frac{1}{(2m+1)(2m+2)} + \frac{1}{(2m+2)(2m+3)}$

$\displaystyle = \frac{m+1}{m(2m+1)} - \frac{1}{m(m+1)} + \frac{1}{(2m+1)(2m+2)} + \frac{1}{(2m+2)(2m+3)}$

$\displaystyle = \frac1m \cdot \frac{(m+1)^2-(2m+1)}{(2m+1)(m+1)} + \frac{1}{2(m+1)}\cdot\frac{2m+3+2m+1}{(2m+1)(2m+3)}$

$\displaystyle = \frac{m^2}{m(2m+1)(m+1)} + \frac{4m+4}{2(m+1)(2m+1)(2m+3)}$

$\displaystyle = \frac{m}{(2m+1)(m+1)} + \frac{2m+2}{(m+1)(2m+1)(2m+3)}$

$\displaystyle = \frac{m(2m+3)+2m+2}{(m+1)(2m+1)(2m+3)}$

$\displaystyle = \frac{2m^2+5m+2}{(m+1)(2m+1)(2m+3)}$

$\displaystyle = \frac{(2m+1)(m+2)}{(m+1)(2m+1)(2m+3)}$

$\displaystyle = \frac{(m+2)}{(m+1)(2m+3)} = \frac{m+1+1}{(m+1)(2(m+1)+1)}$

hence, P(m+1) is true. By math induction, $P(n) \text{ is true } \forall n \in \mathbb{N}$ .

(b)

$\displaystyle \sum_{k=50}^{200}\frac{1}{k(k+1)} = \sum_{k=50}^{100}\frac{1}{k(k+1)} + \sum_{k=100}^{200}\frac{1}{k(k+1)} - \frac{1}{100(101)}$

$\displaystyle = \frac{51}{50(101)} + \frac{101}{100(201)} - \frac{1}{100(101)}$

$\displaystyle = \frac{102}{100(101)} + \frac{101}{100(201)} - \frac{1}{100(101)}$

$\displaystyle = \frac{101}{100(101)} + \frac{101}{100(201)}$

$\displaystyle = \frac{101}{100}\left(\frac{1}{101} + \frac{1}{201}\right)$

$\displaystyle = \frac{101}{100}\cdot \frac{302}{101\cdot201}$

$\displaystyle = \frac{302}{20100}$

$\displaystyle = \frac{151}{10050}$

DSE_M2_2019_Q6

(a) (i)

Let $\displaystyle A=\begin{pmatrix} 1 & -2 & -2 \\ 5 & \alpha & \alpha \\ 7 & \alpha-3 & 2\alpha+1 \end{pmatrix} \qquad b=\begin{pmatrix}\beta \\ 5\beta \\ 8\beta \end{pmatrix}$

The system (E) has solution when $\det(A) \neq 0$

$\displaystyle \det A = \begin{vmatrix} 1 & -2 & -2 \\ 5 & \alpha & \alpha \\ 7 & \alpha-3 & 2\alpha+1 \end{vmatrix} = \begin{vmatrix} 1 & -2 & 0 \\ 5 & \alpha & 0 \\ 7 & \alpha-3 & \alpha+4 \end{vmatrix} = (\alpha + 4)(\alpha + 10)$

$\alpha \neq -4 \quad\text{and}\quad \alpha \neq -10$

(a) (ii)

Apply Cramer’s Rule,

$\displaystyle D_y = \begin{vmatrix} 1 & \beta & -2 \\ 5 & 5\beta & \alpha \\ 7 & 8\beta & 2\alpha+1 \end{vmatrix} = \begin{vmatrix} 1 & 0 & -2 \\ 5 & 0 & \alpha \\ 7 & \beta & 2\alpha+1 \end{vmatrix} = -\beta(\alpha+10)$

So,

$\displaystyle y = \frac{D_y}{\det A} = \frac{-\beta(\alpha+10)}{(\alpha+4)(\alpha+10)} = -\frac{\beta}{\alpha+4}$

(b) If $\alpha = -4$ , then $A^{-1}$ doesn’t exist. So the only way that Ax = b having a solution is when b is a zero vector or $\beta = 0$ .

Now, we want the system (E) to be inconsistent, so $\beta \neq 0$ .

DSE_M2_2019_Q7

(a)

$\displaystyle \int e^x\sin \pi x \text{ d}x = \int \sin \pi x \text{ d}e^x$

$\displaystyle = e^x \sin \pi x - \pi \int e^x \cos \pi x \text{ d}x$

$\displaystyle = e^x \sin \pi x - \pi \int \cos \pi x \text{ d}e^x$

$\displaystyle = e^x \sin \pi x - \pi e^x \cos \pi x - \pi^2 \int \cos \pi x \text{ d}e^x$

$\displaystyle = \frac{e^x \sin \pi x - \pi e^x \cos \pi x}{1+\pi^2} + C$

(b)

$\displaystyle \int_0^3 e^{3-x} \sin \pi x \text{ d}x$

Let $t = 3-x \qquad -\text{d}t = \text{d}x$

$\displaystyle = \int_0^3 e^t \sin \pi(3-t) \text{ d}t$

$\displaystyle = \int_0^3 e^t \sin (3\pi-\pi t) \text{ d}t$

$\displaystyle = \int_0^3 e^t (\sin 3\pi \cos \pi t - \cos 3\pi \sin \pi t ) \text{ d}t$

$\displaystyle = \int_0^3 e^t \sin \pi t \text{ d}t$

$\displaystyle = \frac{e^t \sin \pi t - \pi e^t \cos \pi t}{1+\pi^2} \bigg|_0^3$

$\displaystyle = \frac{1}{1+\pi^2} (-\pi e^3(-1)-(0-\pi))$

$\displaystyle = \frac{\pi e^3 + \pi}{1+\pi^2}$

DSE_M2_2019_Q8

(a)

$\displaystyle h'(x) = \frac{2x^2-7x+8}{x} = \frac{2(x^2-\frac72 x)+8}{x} = \frac{1}{x}\left(2\left(x^2-\frac72x+\frac{49}{16}-\frac{49}{16}\right)+8\right)$

$\displaystyle = \frac{1}{x}\left(2\left(x-\frac72\right)^2-\frac{49}{8}+\frac{64}{8}\right) = \frac1x\left(2\left(x-\frac72\right)^2+\frac{15}{8}\right) > 0 \quad \text{since } x>0$

Therefore, h(x) is an increasing function for x > 0.

(b) (i)

$\displaystyle h(x) = \int \frac{2x^2-7x+8}{x} \text{ d}x = \int 2x - 7 + \frac8x \text{ d}x$

$\displaystyle = x^2 - 7x + 8 \ln |x| + C = x^2 - 7x + 8 \ln x + C \quad \text{since } x>0$

To find C, plug in the point (1, 3).

$1^2 - 7 + 0 + C = 3 \quad\Rightarrow\quad C=9$

$\therefore h(x) = x^2-7x+8\ln x +9$

(b) (ii)

$\displaystyle h''(x) = \left(\frac{2x^2-7x+8}{x}\right)' = \left(2x-7+\frac8x\right)' = 2 - \frac{8}{x^2}$

Let $h''(x)=0$ ,

$\displaystyle 2-\frac{8}{x^2}=0$

$\displaystyle 2=\frac{8}{x^2}$

$\displaystyle x^2=4$

$\displaystyle x=2 \quad (\text{reject }x=-2)$

$h(2) = 2^2-7(2)+8\ln 2 + 9 = 8\ln 2 -1$

To show that this is an inflection point:

For 0 < x < 2,

$\displaystyle x^2<4 \Leftrightarrow 2x^2<8 \Leftrightarrow 2<\frac{8}{x^2} \Leftrightarrow h''(x) = 2-\frac{8}{x^2}<0$

For x > 2,

$\displaystyle x^2>4 \Leftrightarrow 2x^2>8 \Leftrightarrow 2>\frac{8}{x^2} \Leftrightarrow h''(x) = 2-\frac{8}{x^2}>0$

Therefore, the inflection point is $(2, 8 \ln 2 - 1)$ .

DSE_M2_2019_Q9

(a) Equation of L:

$\displaystyle y'=\frac{-2x}{6\sqrt{12-x^2}} = -\frac{x}{3\sqrt{12-x^2}}$

$\displaystyle y'(3) = -\frac{3}{3\sqrt3} = -\frac1{\sqrt3}$

$\displaystyle y(3) = \frac{1}{3} \sqrt{12-3^2} = \frac{\sqrt3}{3}$

Then plug them into the point slope form of the equation.

$\displaystyle y-\frac{\sqrt3}{3} = -\frac{1}{\sqrt3}(x-3)$

$\displaystyle \sqrt3 y-1 = -(x-3)$

$x + \sqrt3 y - 4 = 0$

(b) (i)

$\displaystyle L: \quad y = \frac{4-x}{\sqrt3}$

$\displaystyle \frac{4-x}{\sqrt3} = \sqrt{4-x^2}$

$\displaystyle \frac{(4-x)^2}{3} = 4-x^2$

$16-8x+x^2=12-3x^2$

$4x^2-8x+4=0$

$x^2-2x+1=0$

$x=1$

$y=\sqrt{4-1^2}=\sqrt3$

So, point of contact = $(1, \sqrt3)$

(b) (ii) Intersection of C and $\Gamma$ .

$\displaystyle \sqrt{4-x^2} = \frac13\sqrt{12-x^2}$

$\displaystyle 9(4-x^2) = 12-x^2$

$\displaystyle 36-9x^2=12-x^2$

$\displaystyle 8x^2=24$

$\displaystyle x=\sqrt3 \quad (\text{reject }x=-\sqrt3) \text{ since } 0<x<2$

$y=\sqrt{4-3}=1$

Therefore, intersection of C and $\Gamma = (\sqrt3,1)$ .

(b) (iii)

$\displaystyle \text{Area} = \int_1^{\sqrt3} \frac{4-x}{\sqrt3} - \sqrt{4-x^2} \text{ d}x + \int_{\sqrt3}^3 \frac{4-x}{\sqrt3}-\frac{1}{3}\sqrt{12-x^2} \text{ d}x$

$\displaystyle = \frac{1}{\sqrt3}\int_1^3 (4-x)\text{ d}x - \int_1^{\sqrt3} \sqrt{4-x^2} \text{ d}x - \frac13\int_{\sqrt3}^3 \sqrt{12-x^2} \text{ d}x$

$\text{(Let } x=2\sin\theta \text{ in the second integral and } x=\sqrt{12}\sin\theta \text{ in the third integral)}$

$\displaystyle = \frac{1}{\sqrt3}\left(4x-\frac{x^2}{2}\right)\bigg|_1^3 - 4\int_{\pi/6}^{\pi/3} \cos^2\theta \text{ d}\theta - \frac{1}{3}\cdot12 \int_{\pi/6}^{\pi/3} \cos^2\theta \text{ d}\theta$

$\displaystyle = \frac{1}{\sqrt3}\left(4(3)-\frac{3^2}{2}-4(1)+\frac{1}{2}\right) - 8 \int_{\pi/6}^{\pi/3} \cos^2 \theta \text{ d}\theta$

$\displaystyle = \frac{4}{\sqrt3} - 8\int_{\pi/6}^{\pi/3} \frac{1+\cos 2\theta}{2} \text{ d}\theta$

$\displaystyle = \frac{4}{\sqrt3} - 4\left(\theta+\frac{\sin 2\theta}{2}\right)\bigg|_{\pi/6}^{\pi/3}$

$\displaystyle = \frac{4}{\sqrt3} - 4\left(\frac{\pi}{3}+\frac{\sqrt3}{4}-\frac{\pi}{6}-\frac{\sqrt3}{4}\right)$

$\displaystyle = \frac{4\sqrt3}{3} - \frac{2\pi}{3} = \frac{4\sqrt3-2\pi}{3}$

DSE_M2_2019_Q10

(a)

$\displaystyle \frac{1}{2+\cos 2x} = \frac{1}{2+\cos^2 x - \sin^2 x} = \frac{\sec^2 x}{2\sec^2 x + 1 - \tan^2 x} = \frac{\sec^2 x}{2 \sec^2 x + 1 - (\sec^2 x - 1)}$

$\displaystyle = \frac{\sec^2 x}{2 + \sec^2 x}$

(b)

$\displaystyle \int_0^{\pi/4} \frac{1}{2+\cos 2x}\text{ d}x = \int_0^{\pi/4} \frac{\sec^2 x}{2+\sec^2 x}\text{ d}x$

$\text{Let } w = \tan x$

$\displaystyle = \int_0^1 \frac{\text{d}w}{3+w^2} = \frac13 \int_0^1 \frac{\text{d}w}{1+(w/\sqrt3)^2} = \frac{1}{\sqrt3} \tan^{-1} \frac{w}{\sqrt3} \bigg|_0^1 = \frac{1}{\sqrt3}\cdot\frac{\pi}{6} = \frac{\pi\sqrt3}{18}$

(c)

$\displaystyle \int_{-a}^a f(x) \ln(1+e^x) \text{ d}x = \int_{-a}^0 f(x) \ln(1+e^x)\text{ d}x + \int_0^a f(x) \ln(1+e^x) \text{ d}x$

$\text{Let } x = -w$

$\displaystyle = \int_0^a f(-w) \ln(1+e^{-w}) \text{ d}w + \int_0^a f(x) \ln(1+e^x) \text{ d}x$

$\displaystyle = -\int_0^a f(x) \ln \frac{e^x+1}{e^x} \text{ d}x + \int_0^a f(x) \ln(1+e^x) \text{ d}x$

$\displaystyle = -\int_0^a f(x) \ln (e^x+1) \text{ d}x + \int_0^a f(x) \ln e^x \text{ d}x + \int_0^a f(x) \ln(1+e^x) \text{ d}x$

$\displaystyle = \int_0^a f(x) \ln e^x \text{ d}x = \int_0^a x f(x) \text{ d}x$

(d)

Since $\displaystyle \frac{\sin 2(-x)}{(2+\cos 2(-x))^2} = -\frac{\sin 2x}{(2+\cos 2x)^2}$ , then by (c),

$\displaystyle \int_{-\pi/4}^{\pi/4} \frac{\sin 2x}{(2+\cos 2x)^2} \ln(1+e^x) \text{ d}x = \int_0^{\pi/4} x \cdot \frac{\sin 2x}{(2+\cos 2x)^2} \text{ d}x = \frac12 \int_0^{\pi/4} x \text{ d}\left(\frac{1}{2+\cos2x}\right)$

$\displaystyle = \frac12 \left(\frac{x}{2+\cos2x}\bigg|_0^{\pi/4} - \int_0^{\pi/4}\frac{1}{2+\cos2x}\text{ d}x \right) = \frac12\left(\frac{\pi/4}{2+0}-\frac{\pi\sqrt3}{18}\right)$

$\displaystyle = \frac12\left(\frac{\pi}{8}-\frac{\pi\sqrt3}{18}\right) = \frac{\pi}{16} - \frac{\pi\sqrt3}{36}$

DSE_M2_2019_Q11

(a)

$\displaystyle M^2 = \begin{pmatrix} 2 & 7 \\ -1 & -6 \end{pmatrix} \begin{pmatrix} 2 & 7 \\ -1 & -6 \end{pmatrix} = \begin{pmatrix} -3 & -28 \\ 4 & 29 \end{pmatrix} = -4\begin{pmatrix} 2 & 7 \\ -1 & -6 \end{pmatrix} + 5\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$

$\displaystyle \therefore a = -4, b=5$

(b)

$\displaystyle \text{Let } P(n): \quad 6M^n=(1-(-5)^n)M+(5+(-5)^n)I \quad \forall n\in\mathbb{N}$

When n = 1,

$LS = 6M$

$RS = (1-(-5))M + (5+(-5))I = 6M = LS$

P(1) is true.

$\displaystyle \text{Assume } P(k): \quad 6M^k=(1-(-5)^k)M+(5+(-5)^k)I$

P(k+1):

$\displaystyle 6M^{k+1} = 6M^k\cdot M = \left(\left(1-(-5)^k\right)M+\left(5+(-5)^k\right)I\right)\cdot M$

$\displaystyle = \left(1-(-5)^k\right)M^2+\left(5+(-5)^k\right)M$

By (a)

$\displaystyle = \left(1-(-5)^k\right)(-4M+5I)+\left(5+(-5)^k\right)M$

$\displaystyle =\left(-4+4(-5)^k+5+(-5)^k\right)M + \left(5-5(-5)^k\right)I$

$\displaystyle =\left(1+5(-5)^k\right)M + \left(5-5(-5)^k\right)I$

$\displaystyle =\left(1-(-5)^{k+1}\right)M + \left(5+(-5)^{k+1}\right)I$

hence P(k + 1) is true.

By Math Induction, P(n) is true for all positive integers n.

(c)

In (b),

$\displaystyle 6M^n=(1-(-5)^n)M+(5+(-5)^n)I \quad \forall n\in\mathbb{N}$

$\displaystyle M^n = \frac{1-(-5)^n}{6}M + \frac{5+(-5)^n}{6}I$

$\displaystyle =\frac16 \begin{pmatrix} 2(1-(-5)^n)+5+(-5)^n & 7(1-(-5)^n) \\ -(1-(-5)^n) & -6(1-(-5)^n)+5+(-5)^n \end{pmatrix}$

$\displaystyle =\frac16 \begin{pmatrix} 7-(-5)^n & 7-7(-5)^n) \\ -1+(-5)^n) & -1+7(-5)^n \end{pmatrix}$

$\displaystyle \det M^n = (\det M)^n = (-12+7)^n = (-5)^n$

$\displaystyle (M^n)^{-1} = \frac{1}{6(-5)^n} \begin{pmatrix} -1+7(-5)^n & -7+7(-5)^n \\ 1-(-5)^n & 7-(-5)^n \end{pmatrix} = \frac{1}{6(-5)^n} \left( \begin{pmatrix} -1 & -7 \\ 1 & 7 \end{pmatrix} + (-5)^n \begin{pmatrix} 7 & 7 \\ -1 & -1 \end{pmatrix} \right)$

$\displaystyle =\frac16\begin{pmatrix} 7 & 7 \\ -1 & -1 \end{pmatrix} + \frac{1}{(-5)^n} \cdot \frac16 \begin{pmatrix} -1 & -7 \\ 1 & 7 \end{pmatrix}$

Therefore matrices A and B do exist and they are:

$\displaystyle A = \begin{pmatrix} 7/6 & 7/6 \\ -1/6 & -1/6 \end{pmatrix} \quad \text{and}\quad B = \begin{pmatrix} -1/6 & -7/6 \\ 1/6 & 7/6 \end{pmatrix}$

DSE_M2_2019_Q12

(a)

$\left|\overrightarrow{AC}\right| = \left|\overrightarrow{OC} - \overrightarrow{OA}\right| = \left|-5\mathbf{i} - 12\mathbf{j} + t\mathbf{k} - \mathbf{i} + 4\mathbf{j} - 2\mathbf{k}\right| = \left|-6\mathbf{i} - 8\mathbf{j} + (t-2)\mathbf{k}\right|$

$\left|\overrightarrow{BC}\right| = \left|\overrightarrow{OC} - \overrightarrow{OB}\right| = \left|-5\mathbf{i} - 12\mathbf{j} + t\mathbf{k} +5 \mathbf{i} + 4\mathbf{j} - 8\mathbf{k}\right| = \left|-8\mathbf{j} + (t-8)\mathbf{k}\right|$

$\displaystyle \left|\overrightarrow{AC}\right| = \left|\overrightarrow{BC}\right|$

$\displaystyle 36+64+(t-2)^2=64+(t-8)^2$

$\displaystyle 36+t^2-4t+4=t^2-16t+64$

$\displaystyle 12t = 24$

$\displaystyle t=2$

(b)

$\displaystyle \overrightarrow{AB} = \overrightarrow{OB}-\overrightarrow{OA} = -5\mathbf{i}-4\mathbf{j}+8\mathbf{k}-1\mathbf{i}+4\mathbf{j}-2\mathbf{k} = -6\mathbf{i}+6\mathbf{k}$

$\displaystyle \overrightarrow{AC} = -6\mathbf{i} - 8\mathbf{j}$

$\displaystyle \overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -6 & 0 & 6 \\ -6 & -8 & 0 \end{vmatrix} = 48\mathbf{i} - 36\mathbf{j} + 48\mathbf{k}$

(c)

$\displaystyle \text{Volume} = \frac16 \left|\overrightarrow{OA} \cdot (\overrightarrow{AB} \times \overrightarrow{AC})\right| = \frac16\left| \left(\mathbf{i}-4\mathbf{j}+2\mathbf{k}\right) \cdot \left(48\mathbf{i}-36\mathbf{j}+48\mathbf{k}\right)\right|$

$\displaystyle =\frac16 \left|48+144+96\right| = 8 + 24 + 16 = 48$

(d) (i)

Since point P is co-planar with points A, B, and C. So, $\overrightarrow{AP} \cdot \left(\overrightarrow{AB} \times \overrightarrow{AC}\right) = 0$

$\overrightarrow{AP} = \overrightarrow{OP} - \overrightarrow{OA} = p\mathbf{i} - \mathbf{i} + 4\mathbf{j} - 2\mathbf{k} = (p-1)\mathbf{i} + 4\mathbf{j} - 2\mathbf{k}$

$\left((p-1)\mathbf{i} + 4\mathbf{j} - 2\mathbf{k}\right) \cdot \left(48\mathbf{i} - 36\mathbf{j} + 48\mathbf{k}\right) = 0$

$48(p-1) - 144 - 96 = 0$

$p-1 - 3 - 2 = 0$

$p=6$

Similarly for $\overrightarrow{AQ}$ and $\overrightarrow{AR}$ :

$\overrightarrow{AQ} = \overrightarrow{OQ} - \overrightarrow{OA} = q\mathbf{j} - \mathbf{i} + 4\mathbf{j} - 2\mathbf{k} = -\mathbf{i} + (q+4)\mathbf{j} - 2\mathbf{k}$

$\overrightarrow{AR} = \overrightarrow{OR} - \overrightarrow{OA} = r\mathbf{k} - \mathbf{i} + 4\mathbf{j} - 2\mathbf{k} = -\mathbf{i} + 4\mathbf{j} + (r-2)\mathbf{k}$

$\overrightarrow{AP} \cdot \left(\overrightarrow{AB} \times \overrightarrow{AC}\right) = -48-36(q+4)-2(48) = 0$

$36(q+4) + 3(48) = 0$

$q + 4 + 4 = 0$

$q = -8$

$\overrightarrow{AR} \cdot \left(\overrightarrow{AB} \times \overrightarrow{AC}\right) = -48-36(4)+(r-2)(48) = 0$

$-1-3+r-2=0$

$r=6$

$\therefore pqr = 6(-8)(6) \neq 0$

(d) {ii}

Let $\vec{n}$ be the normal vector to the plane $Pi$ .

$\vec{n} = \overrightarrow{AB} \times \overrightarrow{AB} = 48\mathbf{i} - 36\mathbf{j} + 48\mathbf{k}$

$\displaystyle \hat{n} = \frac{48\mathbf{i} - 36\mathbf{j} + 48\mathbf{k}}{\sqrt{48^2+36^2+48^2}} = \frac{12(4\mathbf{i} - 3\mathbf{j} + 4\mathbf{k})}{12\sqrt{4^2 + 3^2 + 4^2}} = \frac{4\mathbf{i} - 3\mathbf{j} + 4\mathbf{k}}{\sqrt{41}}$

The vector of OD is basically a projection vector of the OA vector onto normal vector n.

$\displaystyle \overrightarrow{OD} = \left(\overrightarrow{OA} \cdot \hat{n}\right) \hat{n}$

$\displaystyle = \left((\mathbf{i}-4\mathbf{j}+2\mathbf{k})\cdot(\frac{4\mathbf{i}-3\mathbf{j}+4\mathbf{k}}{\sqrt{41}})\right) \frac{4\mathbf{i}-3\mathbf{j}+4\mathbf{k}}{\sqrt{41}}$

$\displaystyle = \frac{24}{41} \left(4\mathbf{i}-3\mathbf{j}+4\mathbf{k}\right) = \frac{96}{41}\mathbf{i} - \frac{72}{41}\mathbf{j} + \frac{96}{41}\mathbf{k}$

(d) (iii)

$\displaystyle \overrightarrow{OE} = \frac{1}{p}\mathbf{i} + \frac{1}{q}\mathbf{j} + \frac{1}{r}\mathbf{k} = \frac{1}{6}\mathbf{i} - \frac{1}{8}\mathbf{j} + \frac{1}{6}\mathbf{k} = \frac{1}{24}\left(4\mathbf{i}-3\mathbf{j}+4\mathbf{k}\right)$

$\displaystyle = \frac{41}{24^2}\cdot\frac{24}{41}\left(4\mathbf{i}-3\mathbf{j}+4\mathbf{k}\right) = \frac{41}{24^2}\overrightarrow{OD}$

hence, $\overrightarrow{OD} // \overrightarrow{OE}$ and therefore, the three points D, E, and O are collinear.

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author: kelvinella

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