Week 1

Tutorial 1: Vector Analysis

1. Given vectors with
$P_1=(3, 2, -2)$, $P_2 = (-1, 0, -2)$, and $P_3 = (-2, 1, 5)$
Solve the components and length of the vector for:
a. $P_2-P_1$ and $P_3-P_1$
b. $|3P_1+2P_3|$
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Solution
Question (a)
$P_2-P_1=(-1,0,-2)-(3,2,-2)=\underline{\underline{(-4,-2,0)}}$
$|P_2-P_1|=\sqrt{(-4)^2+(-2)^2+0^2}=\underline{\underline{\sqrt{20}}}$
$P_3-P_1=(-2,1,5)-(3,2,-2)=\underline{\underline{(-5,-1,7)}}$
$|P_3-P_1|=\sqrt{(-5)^2+(-1)^2+(7)^2}=\sqrt{75}=\underline{\underline{5\sqrt{3}}}$
Question (b)
$\begin{aligned}|3P_1+2P_3|&=|3(3,2,-2)+2(-2,1,5)|\\&=|(9,6,-6)+(-4,2,10)|\\&=|(5,8,4)|\\&=\sqrt{5^2+8^2+4^2}\\&=\underline{\underline{\sqrt{105}}}\end{aligned}$

2. Find $u\cdot v$ and the angle between vector $u$ and $v$ for
a. $u = 2𝑖 − 2𝑗 + 𝑘$, $𝑣 = 3𝑖 + 4𝑘$
b. $u=\sqrt{3}𝑖 − 7𝑗$, $𝑣 =\sqrt{3}𝑖 + 𝑗 − 2𝑘$
c. $𝑢=2𝑖+𝑗$, $𝑣=𝑖+2𝑗−𝑘$
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Solution
Question (a)
$\begin{aligned}\theta&=\cos^{-1}\left(\frac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{u}||\mathbf{v}|}\right)\\&=\cos^{-1}\left(\frac{(2)(3)+(-2)(0)+(1)(4)}{\sqrt{2^2+(-2)^2+1^2}\sqrt{3^2+0^2+4^2}}\right)\\&=\cos^{-1}\left(\frac{10}{\sqrt{9}\sqrt{25}}\right)\\&=\cos^{-1}\left(\frac{2}{3}\right)\\&\approx\underline{\underline{0.84\text{ rad}}}\end{aligned}$
Question (b)
$\begin{aligned}\theta&=\cos^{-1}\left(\frac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{u}||\mathbf{v}|}\right)\\&=\cos^{-1}\left(\frac{(\sqrt{3})(\sqrt{3})+(-7)(1)+(0)(-2)}{\sqrt{(\sqrt{3})^2+(-7)^2+0^2}\sqrt{(\sqrt{3})^2+1^2+(-2)^2}}\right)\\&=\cos^{-1}\left(\frac{3-7}{\sqrt{52}\sqrt{8}}\right)\\&=\cos^{-1}\left(\frac{-1}{\sqrt{26}}\right)\\&\approx\underline{\underline{1.77\text{ rad}}}\end{aligned}$
Question (c)
$\begin{aligned}\theta&=\cos^{-1}\left(\frac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{u}||\mathbf{v}|}\right)\\&=\cos^{-1}\left(\frac{(2)(1)+(1)(2)+(0)(-1)}{\sqrt{2^2+1^2+0^2}\sqrt{1^2+2^2+(-1)^2}}\right)\\&=\cos^{-1}\left(\frac{4}{\sqrt{5}\sqrt{6}}\right)\\&=\cos^{-1}\left(\frac{4}{\sqrt{30}}\right)\\&\approx\underline{\underline{0.75\text{ rad}}}\end{aligned}$

3. Find the area of the triangle determined by the points $P$, $Q$, and $R$ and a unit vector perpendicular to plane $PQR$ for
a. $P (1, -1, 2)$, $Q (2, 0, 1)$, $R (0, 2, 1)$
b. $P (2, -2, 1)$, $Q (3, -1, 2)$, $R (3, -1, 1)$
c. $P (-2, 2, 0)$, $Q (0, 1, -1)$, $R (-1, 2, -2)$
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Solution
Question (a)
$\overrightarrow{PQ}=(2,0,1)-(1,-1,2)=(1,1,-1)$
$\overrightarrow{PR}=(0,2,1)-(1,-1,2)=(-1,3,-1)$
$\overrightarrow{PQ}\times\overrightarrow{PR}=\left|\begin{array}{ccc}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&1&-1\\-1&3&-1\end{array}\right|=2\mathbf{i}+2\mathbf{j}+4\mathbf{k}$
$\text{Area}=\frac{1}{2}\left|\overrightarrow{PQ}\times\overrightarrow{PR}\right|=\frac{1}{2}\sqrt{4+4+16}=\underline{\underline{\sqrt{6}}}$
$\mathbf{u}=\frac{\overrightarrow{PQ}\times\overrightarrow{PR}}{\left|\overrightarrow{PQ}\times\overrightarrow{PR}\right|}=\underline{\underline{\frac{1}{\sqrt{6}}(\mathbf{i}+\mathbf{j}+2\mathbf{k})}}$
Question (b)
$\overrightarrow{PQ}=(3,-1,2)-(2,-2,1)=(1,1,1)$
$\overrightarrow{PR}=(3,-1,1)-(2,-2,1)=(1,1,0)$
$\overrightarrow{PQ}\times\overrightarrow{PR}=\left|\begin{array}{ccc}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&1&1\\1&1&0\end{array}\right|=-\mathbf{i}+\mathbf{j}$
$\text{Area}=\frac{1}{2}\left|\overrightarrow{PQ}\times\overrightarrow{PR}\right|=\frac{1}{2}\sqrt{1+1}=\underline{\underline{\frac{\sqrt{2}}{2}}}$
$\mathbf{u}=\frac{\overrightarrow{PQ}\times\overrightarrow{PR}}{\left|\overrightarrow{PQ}\times\overrightarrow{PR}\right|}=-\frac{1}{\sqrt{2}}(-\mathbf{i}+\mathbf{j})=\underline{\underline{-\frac{1}{\sqrt{2}}(\mathbf{i}-\mathbf{j})}}$
Question (c)
$\overrightarrow{PQ}=(0,1,-1)-(-2,2,0)=(2,-1,-1)$
$\overrightarrow{PR}=(-1,2,-2)-(-2,2,0)=(1,0,-2)$
$\overrightarrow{PQ}\times\overrightarrow{PR}=\left|\begin{array}{ccc}\mathbf{i}&\mathbf{j}&\mathbf{k}\\2&-1&-1\\1&0&-2\end{array}\right|=2\mathbf{i}+3\mathbf{j}+\mathbf{k}$
$\text{Area}=\frac{1}{2}\left|\overrightarrow{PQ}\times\overrightarrow{PR}\right|=\frac{1}{2}\sqrt{4+9+1}=\underline{\underline{\frac{\sqrt{14}}{2}}}$
$\mathbf{u}=\frac{\overrightarrow{PQ}\times\overrightarrow{PR}}{\left|\overrightarrow{PQ}\times\overrightarrow{PR}\right|}=\underline{\underline{\frac{1}{\sqrt{14}}(2\mathbf{i}+3\mathbf{j}+\mathbf{k})}}$

4. Find directional derivative of
a. $f(x,y,z)=x^2yz$ in the direction of $4\hat{i}-3\hat{k}$ at point $(1,-1,1)$.
b. $f(x,y,z)=\sqrt{xyz}$ in the direction of $\hat{v}=(4,-2,2)$ at point $(3,2,6)$.
c. $h(x,y,z)=xy+yz+zx$ in the direction of $\hat{v}=(3,6,-2)$ at point $(1,-1,2)$.
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Solution
Question (a)
$D_uf(x,y,z)=\nabla f(x,y,z)\cdot \mathbf{u}$
$\begin{aligned}\nabla f(x,y,z)&=2xyz\hat{i}+x^2z\hat{j}+x^2y\hat{k}\\\nabla f(1,-1,1)&=-2\hat{i}+\hat{j}-\hat{k}\end{aligned}$
$\mathbf{u}=\frac{\overrightarrow{u}}{|u|}=\frac{4\hat{i}-3\hat{k}}{\sqrt{16+9}}=\frac{4\hat{i}-3\hat{j}}{5}$
$\begin{aligned}D_u&=(-2\hat{i}+\hat{j}-\hat{k})\left(\frac{4\hat{i}-3\hat{j}}{5}\right)\\&=(-2)\left(\frac{4}{5}\right)+0+(1)\cdot\left(\frac{3}{5}\right)\\&=-\frac{8}{5}+\frac{3}{5}\\&=\underline{\underline{-1}}\end{aligned}$
Question (b)
$D_uf(x,y,z)=\nabla f(x,y,z)\cdot \mathbf{u}$
$\begin{aligned}\nabla f(x,y,z)&=\frac{yz}{2\sqrt{xyz}}\hat{i}+\frac{xz}{2\sqrt{xyz}}\hat{j}+\frac{xy}{2\sqrt{xyz}}\hat{k}\\\nabla f(3,2,6)&=\frac{12}{12}\hat{i}+\frac{18}{12}\hat{j}+\frac{6}{12}\hat{k}=\hat{i}+\frac{3}{2}\hat{j}+\frac{1}{2}\hat{k}\end{aligned}$
$\mathbf{u}=\frac{\overrightarrow{u}}{|u|}=\frac{4\hat{i}-2\hat{j}+2\hat{k}}{\sqrt{16+4+4}}=\frac{4\hat{i}-2\hat{j}+2\hat{k}}{2\sqrt{6}}$
$\begin{aligned}D_u&=(\hat{i}+\frac{3}{2}\hat{j}+\frac{1}{2}\hat{k})\left(\frac{4\hat{i}-2\hat{j}+2\hat{k}}{2\sqrt{6}}\right)\\&=\frac{1}{2\sqrt{6}}\left[1(4)+\frac{3}{2}(-2)+\frac{1}{2}(2)\right]\\&=\underline{\underline{\frac{\sqrt{6}}{6}}}\end{aligned}$
Question (c)
$D_uf(x,y,z)=\nabla f(x,y,z)\cdot \mathbf{u}$
$\begin{aligned}\nabla f(x,y,z)&=(y+z)\hat{i}+(x+z)\hat{j}+(y+x)\hat{k}\\\nabla f(1,-1,2)&=\hat{i}+3\hat{j}\end{aligned}$
$\mathbf{u}=\frac{\overrightarrow{u}}{|u|}=\frac{3\hat{i}+6\hat{j}-2\hat{k}}{\sqrt{9+36+4}}=\frac{3\hat{i}+6\hat{j}-2\hat{k}}{7}$
$\begin{aligned}D_u&=(\hat{i}+3\hat{j})\left(\frac{3\hat{i}+6\hat{j}-2\hat{k}}{7}\right)\\&=\frac{1}{7}\left[1(3)+3(6)+0(-2)\right]\\&=\underline{\underline{3}}\end{aligned}$

5. Calculate the divergence of the vector fields $F(x, y)$ and $G (x, y)$
a. $F=y^3\mathbf{i}+xy\mathbf{j}$
b. $G=\frac{4y}{x^2}\mathbf{i}+\sin{(y)}\mathbf{j}+3\mathbf{k}$
c. $G=e^x\mathbf{i}+\ln{(xy)}\mathbf{j}+e^{xyz}\mathbf{k}$
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Solution
Question (a)
$\begin{aligned}\nabla\cdot F(x,y)&=\frac{\partial F_1}{\partial x}+\frac{\partial F_2}{\partial y}\\&=\frac{\partial}{\partial x}y^3+\frac{\partial}{\partial y}xy\\&=0+x\\&=\underline{\underline{x}}\end{aligned}$
Question (b)
$\begin{aligned}\partial\cdot G&=\frac{\partial G_1}{\partial x}+\frac{\partial G_2}{\partial y}+\frac{\partial G_3}{\partial z}\\&=\frac{\partial}{\partial x}\left(\frac{4y}{x^2}\right)+\frac{\partial}{\partial y}\sin{(y)}+\frac{\partial}{\partial z}3\\&=4y\times\frac{\partial}{\partial x}x^{-2}+\cos{(y)}\\&=4y \times (-2)x^{-2-1}+\cos{(y)}\\&=-8yx^{-3}+\cos{(y)}\\&=\underline{\underline{-\frac{8y}{x^3}+\cos{(y)}}}\end{aligned}$
Question (c)
$\begin{aligned}\nabla\cdot G &=\frac{\partial G_1}{\partial x}+\frac{\partial G_2}{\partial y}+\frac{\partial G_3}{\partial z}\\&=\frac{\partial}{\partial x}e^x+\frac{\partial}{\partial y}\ln{(xy)}+\frac{\partial}{\partial z}e^{xyz}\\&=e^x+\frac{\partial}{\partial y}(\ln{(x)}+\ln{(y)})+e^{xyz}\times\frac{\partial}{\partial z}(xyz)\\&=\underline{\underline{e^x+\frac{1}{y}+xye^{xyz}}}\end{aligned}$

6. Calculate the curl of the following vector fields $F(x, y, z)$.
a. $F=3x^2\mathbf{i}+2z\mathbf{j}-x\mathbf{k}$
b. $F=y^3\mathbf{i}+xy\mathbf{j}-z\mathbf{k}$
c. $F=(1+y+z^2)\mathbf{i}+(e^{xyz})\mathbf{j}-(xyz)\mathbf{k}$
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Solution
Question (a)
$\begin{aligned}\nabla \times F&=\left|\begin{array}{ccc}\vec{i}&\vec{j}&\vec{k}\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\3x^2&2z&-x\end{array}\right|\\&=\left(\frac{\partial(-x)}{\partial y}-\frac{\partial(2z)}{\partial z}\right)\vec{i}-\left(\frac{\partial(-x)}{\partial x}-\frac{\partial(3x^2)}{\partial z}\right)\vec{j}+\left(\frac{\partial(2z)}{\partial x}-\frac{\partial(3x^2)}{\partial y}\right)\vec{k}\\&=(0-2)\vec{i}-(-1-0)\vec{j}+(0+0)\vec{k}\\&=\underline{\underline{-2\vec{i}+\vec{j}}}\end{aligned}$
Question (b)
$\begin{aligned}\nabla\times F&=\left(\frac{\partial F_3}{\partial y}-\frac{\partial F_2}{\partial z}\right)\vec{i}-\left(\frac{\partial F_3}{\partial x}-\frac{\partial F_1}{\partial z}\right)\vec{j}+\left(\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\right)\vec{k}\\&=\left(\frac{\partial (-z)}{\partial y}-\frac{\partial (xy)}{\partial z}\right)\vec{i}-\left(\frac{\partial (-z)}{\partial x}-\frac{\partial (y^3)}{\partial z}\right)\vec{j}+\left(\frac{\partial (xy)}{\partial x}-\frac{\partial (y^3)}{\partial y}\right)\vec{k}\\&=0\vec{i}-0\vec{j}+(y-3y^2)\vec{k}\\&=\underline{\underline{(y-3y^2)\vec{k}}}\end{aligned}$
⇒ the curl vector is in the $k$ direction
 
Question (c)
$\begin{aligned}\nabla\times F&=\left(\frac{\partial(-xyz)}{\partial y}-\frac{\partial (e^{xyz})}{\partial z}\right)\vec{i}+\left(\frac{\partial (1+y+z^2)}{\partial z}-\frac{\partial (-xyz)}{\partial x}\right)\vec{j}+\left(\frac{\partial (e^{xyz})}{\partial z}-\frac{\partial (1+y+z^2)}{\partial y}\right)\vec{k}\\&=\underline{\underline{(-xz-xye^{xyz})\vec{i}+(2z+yz)\vec{j}+(yze^{xyz}-1)\vec{k}}}\end{aligned}$