Tutorial 3

Tutorial 3: Vector Algebra I

Tutorial QuestionTutorial Solution

1. Sketch the two points $P(3,-1,5)$ and $Q(2,1,-1)$ in three-dimensional space and find the distance between the two points.

   Solution

   

   Distance between the two points:

   $$\begin{aligned} |\overrightarrow{P Q}|&=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2+\left(z_2-z_1\right)^2}\\ &=\sqrt{(2-3)^2+(1-(-1))^2+(-1-5)^2}\\ &=\sqrt{41} \end{aligned}$$

2. For the position vector $\underset{\sim}{a}=\langle 2,4\rangle$, compute $3 \underset{\sim}{a}, \frac{1}{2} \underset{\sim}{a}$, and $-\underset{\sim}{2} a$. Sketch all four vectors on the same axis system. Discuss the effects of scalar multiplication on the magnitude and direction of the original vector.

   Solution

   

   • The scalar multiplication affects the magnitude of vectors if the scalar is a negative value. For example, $-2\underset{\sim }{a}$ switch the direction in exactly the opposite direction.
   • The scalar multiplication affects the magnitude of vectors if the scalar is not equal to 1.
   • When scalar $> 1$, the length of the vector is stretched. (magnitude increased)
   • When scalar $< 1$, the length of the vector is shrinked. (magnitude decteased)

3. Two vectors are given as $\overrightarrow{O P}=\underset{\sim}{i}+3 \underset{\sim}{j}-7 \underset{\sim}{k}$ and $\overrightarrow{O Q}=\underset{\sim}{5} i-2 \underset{\sim}{j}+4 \underset{\sim}{k}$

   i. Find the unit vector in the direction of $\overrightarrow{P Q}$

   Solution

   $$\begin{aligned} \overrightarrow{PQ} & =\overrightarrow{OQ} -\overrightarrow{OP}\\ & =\langle 5,-2,4\rangle -\langle 1,3,-7\rangle \\ & =\langle 4,-5,11\rangle \end{aligned}$$

   $$\begin{aligned} \overrightarrow{PQ} & =\frac{\overrightarrow{PQ}}{| \overrightarrow{PQ}| }\\ & =\frac{\langle 4,-5,11\rangle }{\sqrt{( 4)^{2} +( -5)^{2} +( 11)^{2}}}\\ & =\frac{1}{9\sqrt{2}} \langle 4,-5,11\rangle \end{aligned}$$

   ii. Find the direction cosines of $\overrightarrow{P Q}$

   Solution

   Comparing each component, we obtain:

   • For component $\underset{\sim }{i}$, $\cos \alpha =\frac{4}{9\sqrt{2}}$
   • For component $\underset{\sim }{j}$, $\cos \beta =\frac{-5}{9\sqrt{2}}$
   • For component $\underset{\sim }{k}$, $\cos \gamma =\frac{11}{9\sqrt{2}}$

   iii. Find the vector with magnitude of 5 in the direction of $\overrightarrow{Q P}$ in polar form

   Solution

   $$\begin{aligned} \cos \alpha =\frac{4}{9\sqrt{2}} & \Longrightarrow \ \alpha =71.68^\circ \\ \ \cos \beta =\frac{-5}{9\sqrt{2}} & \Longrightarrow \ \beta =113.13^\circ \\ \cos \gamma =\frac{11}{9\sqrt{2}} & \Longrightarrow \ \gamma =30.20^\circ \end{aligned}$$

   Direction $\overrightarrow{PQ}$ is in the direction of its unit vector, $\langle \cos( 71.68^\circ ) ,\cos( 113.13^\circ ) ,\cos( 30.20^\circ ) \rangle$.

   Direction $\overrightarrow{QP}$ is in the direction of its unit vector,

   $$\begin{aligned} &\langle \cos( 180-71.68^\circ ) ,\cos( 180-113.13^\circ ) ,\cos( 180-30.20^\circ ) \rangle \\ &=\langle \cos( 108.32^\circ ) ,\cos( 66.87^\circ ) ,\cos( 149.80^\circ ) \rangle \end{aligned}$$

   Vector of magintude 5 in the direction $\overrightarrow{QP}$ is $5\langle \cos( 108.32^\circ ) ,\cos( 66.87^\circ ) ,\cos( 149.80^\circ ) \rangle$.

4. Two points are given as $A(1,2)$ and $B(3,4)$.

   i. Find the vector equation of line $L$ that is passing through point $A$ and $B$.

   Solution

   Direction of vector is parallel with $\overrightarrow{AB} =\overrightarrow{OB} -\overrightarrow{OA} =\langle 3,4\rangle -\langle 1,2\rangle =\langle 2,2\rangle$

   Vector equation of line $L$ is $\underset{\sim }{r} =\underset{\sim }{a} +t\underset{\sim }{v} =\langle 1,2\rangle +t\langle 2,2\rangle$

   ii. Sketch the line for $t=0: 1: 5$ and indicate its direction and initial point.

   Solution

   $t$	$\underset{\sim }{r}$
   0	$\langle 1,2\rangle$
   1	$\langle 3,4\rangle$
   2	$\langle 5,6\rangle$
   3	$\langle 7,8\rangle$
   4	$\langle 9,10\rangle$
   5	$\langle 11,12\rangle$

   

5. If a unit vector $\vec{a}$ makes an angle of $\pi / 3$ with $i, \pi / 4$ with $j$ and acute angle $\theta$ with $k$, find $\theta$ and the components of $\vec{a}$.

   Solution

   Let $\alpha ,\beta ,\gamma$ to represent angles of $\vec{a}$, and $\vec{a} =a_{1} i+a_{2} j+a_{3} k$.

   $$\begin{aligned} \cos \alpha & =\frac{a_{1}}{| a_{1}| }\\ \cos\left(\frac{\pi }{3}\right) & =\frac{a_{1}}{1}\\ a_{1} & =\frac{1}{2} \end{aligned}$$

   $$\begin{aligned} \cos \beta & =\frac{a_{2}}{| a_{2}| }\\ \cos\left(\frac{\pi }{4}\right) & =\frac{a_{1}}{1}\\ a_{1} & =\frac{1}{\sqrt{2}} \end{aligned}$$

   $$\begin{aligned} \cos \gamma & =\frac{a_{3}}{| a_{3}| }\\ \cos \theta & =\frac{a_{3}}{1}\\ a_{3} & =\cos \theta \end{aligned}$$

   $\therefore \ \vec{a} =\frac{1}{2} i+\frac{1}{\sqrt{2}} j+\cos \theta k$

   $$\begin{aligned} | \vec{a}| & =\sqrt{\left(\frac{1}{2}\right)^{2} +\left(\frac{1}{\sqrt{2}}\right)^{2} +(\cos \theta )^{2}}\\ 1 & =\sqrt{\frac{1}{4} +\frac{1}{2} +\cos^{2} \theta }\\ 1 & =\frac{3}{4} +\cos^{2} \theta \\ \cos^{2} \theta & =\frac{1}{4}\\ \cos \theta & =\pm \frac{1}{2}\\ \theta & =60^\circ \ or\ 120^\circ \end{aligned}$$

   $\therefore \ \vec{a} =\frac{1}{2} i+\frac{1}{\sqrt{2}} j+\frac{1}{2} k\ \quad or\quad \vec{a} =\frac{1}{2} i+\frac{1}{\sqrt{2}} j-\frac{1}{2} k$

6. If $\vec{a}$ is a unit vector and $(\vec{x}-\vec{a}) \cdot(\vec{x}+\vec{a})=8$, find $|\vec{x}|$.

   Solution

   Since $\vec{a}$ is a unit vector, $\vert \vec{a}\vert =1$.

   $$\begin{aligned} (\vec{x} -\vec{a}) \cdot (\vec{x} +\vec{a}) & =8\\ \vec{x} \cdot \vec{x} +\vec{x} \cdot \vec{a} -\vec{a} \cdot \vec{x} -\vec{a} \cdot \vec{a} & =8\\ | \vec{x}| ^{2} -| \vec{a}| ^{2} & =8\\ | \vec{x}| ^{2} -1 & =8\\ | \vec{x}| ^{2} & =9\\ | \vec{x}| & =3 \end{aligned}$$

   Note that magnitude of vector is non-negative.

7. Find the gradient for $f=\left(2 x^{2}+y\right) /\left(x^{2}-y^{2}\right)$.

   Solution

   $$\frac{d}{d x}(f)=\frac{\left(x^2-y^2\right)(4 x)-\left(2 x^2+y\right)(2 x)}{\left(x^2-y^2\right)^2}$$

   $$\frac{d}{d x}(f)=\frac{2 x y(-2 y-1)}{\left(x^2-y^2\right)^2}$$

   and

   $$\frac{d}{d y}(f)=\frac{\left(x^2-y^2\right)(1)-\left(2 x^2+y\right)(-2 y)}{\left(x^2-y^2\right)^2}$$

   $$\frac{d}{d y}(f)=\frac{x^2+4 x^2 y+y^2}{\left(x^2-y^2\right)^2}$$

   Hence

   $$\nabla f=\frac{2 x y(-2 y-1)}{\left(x^2-y^2\right)^2} \boldsymbol{i}+\frac{x^2+4 x^2 y+y^2}{\left(x^2-y^2\right)^2} \boldsymbol{j}$$

8. Calculate the divergence of the following vector fields of $F(x, y)$ and $G(x, y)$;

   a. $F=y^{3} \boldsymbol{i}+x y \boldsymbol{j}$

   Solution

   $$\begin{aligned} \nabla \cdot \boldsymbol{F}(x, y) &=\frac{\partial F_1}{\partial x}+\frac{\partial F_2}{\partial y} \\ &=\frac{\partial}{\partial x} y^3+\frac{\partial}{\partial y} x y=0+x=x \end{aligned}$$

   b. $G=\frac{4 y}{x^{2}} \boldsymbol{i}+\sin (y) \boldsymbol{j}+3 \boldsymbol{k}$

   Solution

   $$\begin{aligned} \nabla \cdot \boldsymbol{G} &=\frac{\partial G_1}{\partial x}+\frac{\partial G_2}{\partial y}+\frac{\partial G_3}{\partial z} \\ &=\frac{\partial}{\partial x}\left(\frac{4 y}{x^2}\right)+\frac{\partial}{\partial y} \sin (y)+\frac{\partial}{\partial z} 3 \\ &=4 y \times \frac{\partial}{\partial x} x^{-2}+\cos (y)=4 y \times(-2) x^{-2-1}+\cos (y) \\ &=-8 y x^{-3}+\cos (y)=-\frac{8 y}{x^3}+\cos (y) \end{aligned}$$

   c. $G=e^{x} \boldsymbol{i}+\ln (x y) \boldsymbol{j}+e^{x y z} \boldsymbol{k}$

   Solution

   $$\begin{aligned} \nabla \cdot \boldsymbol{G} &=\frac{\partial G_1}{\partial x}+\frac{\partial G_2}{\partial y}+\frac{\partial G_3}{\partial z} \\ &=\frac{\partial}{\partial x} \mathrm{e}^x+\frac{\partial}{\partial y} \ln (x y)+\frac{\partial}{\partial z} \mathrm{e}^{x y z} \\ &=\mathrm{e}^x+\frac{\partial}{\partial y}(\ln (x)+\ln (y))+\mathrm{e}^{x y z} \times \frac{\partial}{\partial z}(x y z) \\ &=\mathrm{e}^x+\frac{1}{y}+x y \mathrm{e}^{x y z} \end{aligned}$$

9. Calculate the curl of the following vector fields of $F(x, y, z)$;

   a. $F=3 x^{2} \boldsymbol{i}+2 z \boldsymbol{j}-x \boldsymbol{k}$

   Solution

   $$\begin{aligned} \nabla \times \boldsymbol{F} &=\left|\begin{array}{ccc}i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 3 x^2 & 2 z & -x\end{array}\right| \\ &=\left(\frac{\partial(-x)}{\partial y}-\frac{\partial(2 z)}{\partial z}\right) i-\left(\frac{\partial(-x)}{\partial x}-\frac{\partial\left(3 x^2\right)}{\partial z}\right) j +\left(\frac{\partial(2 z)}{\partial x}-\frac{\partial\left(3 x^2\right)}{\partial y}\right) k \\ &=(0-2) i-(-1-0) j+(0-0) k \\ &=-2 i+j \end{aligned}$$

   b. $F=y^{3} \boldsymbol{i}+x y \boldsymbol{j}-z \boldsymbol{k}$

   Solution

   $$\begin{aligned} \nabla \times \boldsymbol{F}=&\left(\frac{\partial F_3}{\partial y}-\frac{\partial F_2}{\partial z}\right) i-\left(\frac{\partial F_3}{\partial x}-\frac{\partial F_1}{\partial z}\right) j+\left(\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\right) \boldsymbol{k} \\ =&\left(\frac{\partial(-z)}{\partial y}-\frac{\partial(x y)}{\partial z}\right) i-\left(\frac{\partial(-z)}{\partial x}-\frac{\partial\left(y^3\right)}{\partial z}\right) j +\left(\frac{\partial(x y)}{\partial x}-\frac{\partial\left(y^3\right)}{\partial y}\right) \boldsymbol{k} \\ =& 0 i-0 j+\left(y-3 y^2\right) \boldsymbol{k}\\ =&\left(y-3 y^2\right) \boldsymbol{k} \end{aligned}$$

   i.e., the curl vector is in the $k$ direction.

   с. $F=\left(1+y+z^{2}\right) \boldsymbol{i}+\left(e^{x y z}\right) \boldsymbol{j}-(x y z) \boldsymbol{k}$

   Solution

   $$\operatorname{curl}(\mathbf{F})=\nabla \times \mathbf{F}=\left(\frac{\partial}{\partial y}(x y z)-\frac{\partial}{\partial z}\left(e^{x y z}\right)\right) \vec{i}+\left(\frac{\partial}{\partial z}\left(1+y+z^2\right)-\frac{\partial}{\partial x}(x y z)\right) \vec{j}+\left(\frac{\partial}{\partial x}\left(e^{x y z}\right)-\frac{\partial}{\partial y}\left(1+y+z^2\right)\right) \vec{k}$$

   $$\operatorname{curl}(\mathbf{F})=\left(x z-x y e^{x y z}\right) \vec{i}+(2 z-y z) \vec{k}+\left(y z e^{x y z}-1\right) \vec{k}$$

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