Tutorial 4

Tutorial 4: Vector Algebra II

Tutorial QuestionTutorial Solution

1. Two long straight pipes are specified using Cartesian coordinates as follow:

   Pipe A: diameter $0.8$; axis through points $(2,5,3)$ and $(7,10,8)$.

   Pipe B: diameter $1.0$; axis through points $(0,6,3)$ and $(-12,0,9)$.

   Explain if the two pipes require realignment to prevent intersection.

   Solution

   Pipe A and B have axes:

   $$\begin{aligned} \mathbf{r}_{A} &=[ 2,5,3] +\lambda '[ 5,5,5] =[ 2,5,3] +\lambda \frac{[ 1,1,1]}{\sqrt{3}}]\\ \mathbf{r}_{B} &=[ 0,6,3] +\mu '[ -12,-6,6] =[ 0,6,3] +\mu \frac{[ -2,-1,1]}{\sqrt{6}} \end{aligned}$$

   

   (Non-unit) perpendicular to both axes is

   $$\mathbf{p} =\begin{vmatrix} \hat{\mathbf{i}} & \mathbf{\hat{j}} & \mathbf{\hat{k}}\\ 1 & 1 & 1\\ -2 & -1 & 1 \end{vmatrix} =[ 2,-3,1]$$

   The length of the mutual perpendicular is mod

   $(\mathbf{a} -\mathbf{b}) \cdot \frac{[ 2,-3,1]}{\sqrt{14}} =[ 2,-1,0] \cdot \frac{[ 2,-3,1]}{\sqrt{14}} =1.87$

   Sum of the radii of the pipes is $0.4+0.5=0.9$. Hence, the pipes do not intersect.

2. Determine if the sets of vectors given are parallel or non-parallel. Show your answers:

   i. $\underset{\sim}{a}=\langle 2,4,-1\rangle$ and $\underset{\sim}{b}=\langle-6,-12,3\rangle$

   Solution

   These two vectors are parallel since $\underset{\sim}{b}=-3\underset{\sim}{a}$

   ii. $\underset{\sim}{a}=\langle 4,10\rangle$ and $\underset{\sim}{b}=\langle 2,-9\rangle$

   Solution

   These two vectors are not parallel since there is no scalar $a$ that can fulfil the scalar multiplication where $\underset{\sim}{a}\neq a\underset{\sim}{b}$

3. Find the unit vectors that are perpendicular to the vectors $\underset{\sim}{a}$ and $\underset{\sim}{b}$ as following:

   i. $\underset{\sim}{a}=\langle 2,4,5\rangle, \underset{\sim}{\underset{\sim}{b}}=\langle 1,2,-2\rangle$

   Solution

   Vector perpendicular to both vector $\underset{\sim }{a}$ and $\underset{\sim }{b}$ is $\underset{\sim }{a} \times \underset{\sim }{b}$.

   Unit vector of $\underset{\sim }{a} \times \underset{\sim }{b}$ is $\underset{\sim }{a} \times \underset{\sim }{b} =\frac{\underset{\sim }{a} \times \underset{\sim }{b}}{\left\vert \underset{\sim }{a} \times \underset{\sim }{b}\right\vert }$.

   $$\underset{\sim }{a} \times \underset{\sim }{b} =\begin{vmatrix} \underset{\sim }{i} & \underset{\sim }{j} & \underset{\sim }{k}\\ 2 & 4 & 5\\ 1 & 2 & -2 \end{vmatrix} =\underset{\sim }{i}( -8+10) -\underset{\sim }{j}( -4+5) +\underset{\sim }{k}( 4-4) =\langle -2,-1,0\rangle$$

   $\underset{\sim }{a} \times \underset{\sim }{b} =\frac{\underset{\sim }{a} \times \underset{\sim }{b}}{\left\vert \underset{\sim }{a} \times \underset{\sim }{b}\right\vert } =\frac{\langle -2,-1,0\rangle }{\sqrt{( -2)^{2} +( -1)^{2}}} =\langle \frac{-2}{\sqrt{5}} ,\frac{-1}{\sqrt{5}} ,0\rangle$

   The unit vector is exist in this case.

   ii. $\underset{\sim}{a}=\langle 2,4,-4\rangle, \underset{\sim}{b}=\langle 1,2,-2\rangle$

   Solution

   $$\underset{\sim }{a} \times \underset{\sim }{b} =\begin{vmatrix} \underset{\sim }{i} & \underset{\sim }{j} & \underset{\sim }{k}\\ 2 & 4 & 4\\ 1 & 2 & -2 \end{vmatrix} =\underset{\sim }{i}( -8+8) -\underset{\sim }{j}( -4+4) +\underset{\sim }{k}( 4-4) =\langle 0,0,0\rangle$$

   $\underset{\sim }{a} \times \underset{\sim }{b} =\frac{\underset{\sim }{a} \times \underset{\sim }{b}}{\left\vert \underset{\sim }{a} \times \underset{\sim }{b}\right\vert } =\frac{\langle 0,0,0\rangle }{0}$

   The unit vector does not exist in this case.

   We know that the unit vector always have magnitude of 1. However since the zero vector has magnitude of 0, the unit vector of zero vector does not exist. In common practice, we have several remedy options depending on your application:

   • Return $\underset{\sim }{a}\times\underset{\sim }{b}=$ zero vector
   • Return $(\underset{\sim }{a}\times\underset{\sim }{b})=NaN$ (i.e, Not a number)

4. Find the distance from $P=(-3,7,4)$ to the line $l$ with vector equation;

   $$r=\left(\begin{array}{c} 2 \\ -2 \\ -3 \end{array}\right)+\lambda\left(\begin{array}{c} 4 \\ -5 \\ 3 \end{array}\right)$$
   Solution

   Here $\mathbf{a} =\begin{pmatrix}2\\-2\\-3\end{pmatrix}$, $A=( 2,-2,3)$ and $\mathbf{u}=\begin{pmatrix}4\\-5\\3\end{pmatrix}$. So, $\overrightarrow{AP}$ represents $\mathbf{v} =\mathbf{p} -\mathbf{a} =\begin{pmatrix}-3\\7\\4\end{pmatrix} -\begin{pmatrix}2\\-2\\-3\end{pmatrix} =\begin{pmatrix}-5\\9\\7\end{pmatrix}$. Now,

   $$\mathbf{u} \times \mathbf{v} =\begin{vmatrix} -5 & 3\\ 9 & 7 \end{vmatrix}\mathbf{i} -\begin{vmatrix} 4 & 3\\ -5 & 7 \end{vmatrix}\mathbf{j} +\begin{vmatrix} 4 & -5\\ -5 & 9 \end{vmatrix}\mathbf{k} =\begin{pmatrix} -62\\ -43\\ 11 \end{pmatrix}$$

   Therefore, $\vert \mathbf{u} \times \mathbf{v}\vert =\sqrt{( -62)^{2} +( -43)^{2} +11^{2}} =3\sqrt{646}$ and $\vert \mathbf{u}\vert =\sqrt{4^{2} +( -5)^{2} +3^{2}} =5\sqrt{2}$.

   The distance is,

   $$\frac{\vert\mathbf{u} \times \mathbf{v}\vert}{\vert \mathbf{u}\vert } =\frac{3\sqrt{646}}{5\sqrt{2}} =\frac{3\sqrt{323}}{5}$$

   Note that we have $\mathbf{v} \times \mathbf{u} =(\mathbf{p} -\mathbf{a}) \times \mathbf{u} =-(\mathbf{u} \times \mathbf{v})$ and so $\vert \mathbf{v} \times \mathbf{u}\vert =\vert (\mathbf{p} -\mathbf{a}) \times \mathbf{u}\vert =\vert \mathbf{u} \times \mathbf{v}\vert$.

5. Calculate the distance between the lines $l$ and $m$ having vector equations $\boldsymbol{r}=\boldsymbol{a}+\lambda \boldsymbol{u}$ and $\boldsymbol{r}=\boldsymbol{b}+\mu \boldsymbol{v}$ respectively, where;

   $$\boldsymbol{a}=\left(\begin{array}{c} 0 \\ 4 \\ -1 \end{array}\right), \boldsymbol{u}=\left(\begin{array}{c} 1 \\ -3 \\ -2 \end{array}\right), \boldsymbol{b}=\left(\begin{array}{c} 2 \\ -1 \\ 0 \end{array}\right) \text { and } \boldsymbol{v}=\left(\begin{array}{c} -3 \\ 1 \\ 2 \end{array}\right)$$
   Solution

   We have $\mathbf{b} -\mathbf{a} =2\mathbf{i} -5\mathbf{j} +k$ and

   $$\mathbf{u} \times \mathbf{v} =\begin{pmatrix} 1\\ -3\\ -2 \end{pmatrix} \times \begin{pmatrix} -3\\ 1\\ 2 \end{pmatrix} =\begin{vmatrix} -3 & 1\\ -2 & 2 \end{vmatrix}\mathbf{i} -\begin{vmatrix} 1 & -3\\ -2 & 2 \end{vmatrix}\mathbf{j} +\begin{vmatrix} 1 & -3\\ -3 & 1 \end{vmatrix}\mathbf{k} =\begin{pmatrix} -4\\ 4\\ -8 \end{pmatrix}$$

   Thus, we get $(\mathbf{b} -\mathbf{a}) \cdot (\mathbf{u} \times \mathbf{v}) =-8-20-8=-36$ and $\vert \mathbf{u} \times \mathbf{v}\vert =\sqrt{( -4)^{2} +4^{2} +( -8)^{2}} =4\sqrt{6}$ The distance from $l$ to $m$ is

   $$\frac{\vert \mathbf{( b - a) \cdot ( u} \times \mathbf{v})\vert }{\vert \mathbf{u} \times \mathbf{v}\vert } =\frac{\vert -36\vert }{\sqrt{96}} =\frac{36}{4\sqrt{6}} =\frac{9}{\sqrt{6}}$$

6. Find the following equation of line for the line $L$ passing through the point $P(3,1,-2)$ and $Q(-2,7,-4)$

   i. Vector equation

   Solution

   $\overrightarrow{PQ}=\overrightarrow{OQ}-\overrightarrow{O P}=\langle-2,7,-4\rangle-\langle 3,1,-2\rangle=\langle-5,6,-2\rangle$ is vector parallel to the line $L$

   The vector equation of the line, $\underset{\sim}{r}=\underset{\sim}{a}+t\underset{\sim}{v}$ where $\underset{\sim}{a}$ is position vector of any point at the line and $\underset{\sim}{v}$ is vector parallel to the line.

   Thus, $\underset{\sim}{r}=\langle 3,1,-2\rangle+t\langle-5,6,-2\rangle$ is the vector equation of line $L$

   Noted that $\underset{\sim}{r}=\langle-2,7,4\rangle+t\langle-5,6,-2\rangle$ is also acceptable. To generate consistent data, we always use the first point.

   ii. Parametric equation

   Solution

   $\underset{\sim}{r}=\langle x, y, z\rangle$

   Compare both side,

   $$\begin{aligned} x&=3-5t\\ y&=1+6t,\quad t\in \mathbb{R}\\ z&=-2-2t \end{aligned}$$

   iii.Cartesian equation

   Solution

   Eliminating the parameter $t$, we get

   $$\frac{x-3}{-5}=\frac{y-1}{6}=\frac{z+2}{-2}$$

7. Find the Cartesian equation of plane contains the point $(1,2,-1)$ and perpendicular to the intersecting line of the planes and $2 x+y+z=2$ and $x+2 y+z=3$.

   Solution

   The intersecting line has vector that is parallel to both planes, which is the cross product of their respectively normal vector,

   $$\underset{\sim }{n}_{1} \times \underset{\sim }{n}_{2} =\begin{vmatrix} \underset{\sim }{i} & \underset{\sim }{j} & \underset{\sim }{k}\\ 2 & 1 & 1\\ 1 & 2 & 1 \end{vmatrix} =\underset{\sim }{i}( 1-2) -\underset{\sim }{j}( 2-1) +\underset{\sim }{k}( 4-1) =\langle -1,-1,3\rangle$$

   The Cartesian equation of plane that perpendicular to the intersecting line has the normal vector that is parallel to the vector $\underset{\sim }{n}_{1} \times \underset{\sim }{n}_{2} =\langle -1,-1,3\rangle$

   Therefore, Cartesian equation of plane is $-x-y+3z=k$. Substitute the point $( 1,2,-1)$ which is located on the plane to find $k$, $k=-1-2+3( -1) =-6$

   $$\therefore -x-y+3z=-6$$

8. Find the Cartesian equation of plane contains the line $L_{1}: \boldsymbol{r}_{\mathbf{1}}=\boldsymbol{a}+t \boldsymbol{u}=\langle 1,-3,4\rangle+$ $\langle 2,1,1\rangle t$ and parallel to the line $L_{2}: \boldsymbol{r}_{2}=\boldsymbol{b}+s \boldsymbol{v}=\langle 0,0,0\rangle+\langle 1,2,3\rangle s$. Then, proof that the plane is parallel to line $L_{2}$ ?

   Solution

   The normal vector of the plane that containing two lines is

   $$\underset{\sim }{u} \times \underset{\sim }{v} =\begin{vmatrix} \underset{\sim }{i} & \underset{\sim }{j} & \underset{\sim }{k}\\ 2 & 1 & 1\\ 1 & 2 & 3 \end{vmatrix} =\underset{\sim }{i}( 3-2) -\underset{\sim }{j}( 6-1) +\underset{\sim }{k}( 4-1) =\langle 1,-5,3\rangle$$

   Therefore, Cartesian equation of plane is $-x-5y+3z=k$. Substitute point $A( 1,-3,4)$ to find $k$,

   $$k=1-5( -3) +3( 4) =28$$

   $$\therefore x-5y+3z=28$$

   Note: Point $B( 0,0,0)$ can't be used to find $k$ because we don't know if it is located on the plane.

   To prove that the plane is parallel to line $L_{2}$, use dot product where dot product of vector normal and parallel to the plane is equal to zero.

   Vector normal to the plane is $\langle 1,-5,3\rangle$, Vector parallel to the plane is $\langle 1,2,3\rangle$.

   Since $\langle 1,-5,3\rangle \cdot \langle 1,2,3\rangle =0$, thus the plane is parallel to line $L_{2}$.

9. Find the Cartesian equation of plane contains the line $L_{1}: \boldsymbol{r}_{\mathbf{1}}=\boldsymbol{a}+t \boldsymbol{u}=\langle-2,3,4\rangle+$ $\langle 1,2,-1\rangle t$ and line $L_{2}: \boldsymbol{r}_{2}=\boldsymbol{b}+s \boldsymbol{v}=\langle 3,4,0\rangle+\langle-1,-2,1\rangle s$.

   Solution

   The plane consists of line $L_{1}$ and $L_{2}$, therefore point $A( -2,3,4)$ and $B( 3,4,0)$ are located on the plane.

   Therefore, $\overrightarrow{AB} =\overrightarrow{OB} -\overrightarrow{OA} =\langle 3,4,0\rangle -\langle -2,3,4\rangle =\langle 5,1,-4\rangle$ is one of the line that is parallel to the plane. (Note that a plane is formed by infinite line in various direction, thus there is more than one line parallel to it)

   Since the plane contains line $L_{1}$ and $L_{2}$, thus the plane is parallel to both of them. From the vector equation of line, the plane is parallel $\underset{\sim }{u} =\langle 1,2,-1\rangle$ and $\underset{\sim }{v} =\langle -1,-2,1\rangle$ respectively. Note that these two lines are parallel to each other. Thus, they lies in the same plane. Use either one of them to find the plane equation.

   Use cross product to find vector normal to the plane,

   $$\overrightarrow{AB} \times \underset{\sim }{u} =\begin{vmatrix} \underset{\sim }{i} & \underset{\sim }{j} & \underset{\sim }{k}\\ 5 & 1 & -4\\ 1 & 2 & -1 \end{vmatrix} =\underset{\sim }{i}( -1+8) -\underset{\sim }{j}( -5+4) +\underset{\sim }{k}( 10-1) =\langle 7,1,9\rangle$$

   Therefore, Cartesian equation is $7x+y+9z=k$.

   Substitute either $A( -2,3,4)$ or $B( 3,4,0)$ to get $$k$,

   $$k=7( -3) +3+9( 4) =25$$

   $$\therefore 7x+y+9z=25$$

10. Let $\underset{\sim}{a}=\langle 1,-2,-3\rangle, \underset{\sim}{b}=\langle 2,1,-1\rangle$ and $\underset{\sim}{c}=\langle 1,3,-2\rangle$. Find

    i. $\underset{\sim}{a} \cdot \underset{\sim}{b}(\underset{\sim}{a} \times \underset{\sim}{b})$

    Solution

    $$\begin{aligned} \underset{\sim }{a} \cdot \underset{\sim }{b}\left(\underset{\sim }{a} \times \underset{\sim }{b}\right) & =\langle 1,-2,-3\rangle \cdot \langle 2,1,-1\rangle ( \langle 1,-2,-3\rangle \times \langle 2,1,-1)\\ & =( 2-2+3)\left(\begin{vmatrix} \underset{\sim }{i} & \underset{\sim }{j} & \underset{\sim }{k}\\ 1 & -2 & -3\\ 2 & 1 & -1 \end{vmatrix}\right)\\ & =3\left[\underset{\sim }{i}( 2+3) -\underset{\sim }{j}( -1+6) +\underset{\sim }{k}( 1+4)\right]\\ & =3\left( 5\underset{\sim }{i} -5\underset{\sim }{j} +5\underset{\sim }{k}\right)\\ & =15\underset{\sim }{i} -15\underset{\sim }{j} +15\underset{\sim }{k} \end{aligned}$$

    ii. $(\underset{\sim}{a}+\underset{\sim}{b}) \times \underset{\sim}{c}$

    Solution

    $$\begin{aligned} \underset{\sim }{a} +\underset{\sim }{b} \times \underset{\sim }{c} & =( \langle 1,-2,-3\rangle +\langle 2,1,-1\rangle ) \times \langle 1,3,-2\rangle \\ & =( \langle 3,-1,-4\rangle ) \times \langle 1,3,-2\rangle \\ & =\begin{vmatrix} \underset{\sim }{i} & \underset{\sim }{j} & \underset{\sim }{k}\\ 3 & -1 & -4\\ 1 & 3 & -2 \end{vmatrix}\\ & =14\underset{\sim }{i} +2\underset{\sim }{j} +10\underset{\sim }{k} \end{aligned}$$

11. Determine the shortest distance between the 3D skew lines where;

    $$\begin{aligned} &L_{a}: r_{a}=4 \boldsymbol{i}+2 \boldsymbol{j}-6 \boldsymbol{k}+t(2 \boldsymbol{i}-\boldsymbol{j}-\boldsymbol{k}) \\ &L_{b}: r_{b}=\boldsymbol{i}-3 \boldsymbol{j}-3 \boldsymbol{k}+t(\boldsymbol{i}-2 \boldsymbol{j}-\boldsymbol{k}) \end{aligned}$$
    Solution

    Shortest distance is given as: $\left|\left(a_a-a_b\right) \cdot \frac{b_a \times b_b}{\left|b_a \times b_b\right|}\right|$

    $$\begin{gathered} L_a: r_a=4 \boldsymbol{i}+2 \boldsymbol{j}-6 \boldsymbol{k}+t(2 \boldsymbol{i}-\boldsymbol{j}-\boldsymbol{k}) \\ L_b: r_b=\boldsymbol{i}-3 \boldsymbol{j}-3 \boldsymbol{k}+t(\boldsymbol{i}-2 \boldsymbol{j}-\boldsymbol{k}) \\ b_a \times b_b=\left|\begin{array}{ccc} \boldsymbol{i} & \boldsymbol{j} & \boldsymbol{k} \\ 2 & -1 & -1 \\ 1 & -2 & -1 \end{array}\right|=(1-2) \boldsymbol{i}-(-2+1) \boldsymbol{j}+(-4+1) \boldsymbol{k}=-\boldsymbol{i}+\boldsymbol{j}-3 \boldsymbol{k} \\ \left|b_a \times b_b\right|=\sqrt{(-1)^2+1^2+(-3)^2}=\sqrt{11} \\ a_a-a_b=(4-1) \boldsymbol{i}+(2+3) \boldsymbol{j}+(-6+3) \boldsymbol{k}=3 \boldsymbol{i}+5 \boldsymbol{j}-3 \boldsymbol{k} \end{gathered}$$

    Hence, the shortest distance is:

    $$\frac{1}{\sqrt{11}}\left|\begin{array}{cc} 3 & -1 \\ 5 & 1 \\ -3 & -3 \end{array}\right|=\frac{1}{\sqrt{11}}|-3+5+9|=\frac{11}{\sqrt{11}}$$

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