Chapter 9, Problem 5CRT

(a)

To determine

To graph:For the $\text{u}$ and $\text{v}$ in coordinate plane, with initial points $\left(0,0\right)$ .

Explanation of Solution

Given information:

The given vectors are $\text{u}=\langle 8,6\rangle$ and $\text{v}=5\text{i}-10\text{j}$ .

Graph:

The vector $\text{v}=5\text{i}-10\text{j}$ can be written as $\text{v}=\langle 5,-10\rangle$ .

For a vector $\text{u}=\langle a,b\rangle$ with initial point $\left(p,q\right)$ , the terminal point is $\left(p+a,q+b\right)$ .

So, For a vector $\text{u}=\langle 8,6\rangle$ with initial point $\left(0,0\right)$ , the terminal point is $\left(0+8,0+6\right)=\left(8,6\right)$ .

The graph for the vector $\text{u}=\langle 8,6\rangle$ with initial point $\left(0,0\right)$ is shown in figure (1).

  

Figure (1)

For a vector $\text{u}=\langle a,b\rangle$ with initial point $\left(p,q\right)$ , the terminal point is $\left(p+a,q+b\right)$ .

So, For a vector $\text{v}=\langle 5,-10\rangle$ with initial point $\left(0,0\right)$ , the terminal point is $\left(0+5,0+\left(-10\right)\right)=\left(5,-10\right)$ .

The graph for the vector $\text{v}=\langle 5,-10\rangle$ with initial point $\left(0,0\right)$ is shown in figure (2).

  

Figure (2)

Interpretation:Graph for the $\text{u}$ and $\text{v}$ in coordinate plane, with initial points $\left(0,0\right)$ is shown in figure (1) and figure (2) respectively.

(b)

To determine

To find:The value of $\text{u}+\text{v},2\text{u}-\text{v}$ , the angle between $\text{u}$ and $\text{v}$ , and ${\text{proj}}_{\text{v}}\text{u}$ .

Answer to Problem 5CRT

The value of is $13\text{i}-4\text{j}$ , $2\text{u}-\text{v}$ is $11\text{i}+22\text{j}$ , the angle between $\text{u}$ and $\text{v}$ is $100.30°$ , and ${\text{proj}}_{\text{v}}\text{u}$ is $\langle -\frac{4}{5},\frac{8}{5}\rangle$ .

Explanation of Solution

Given information:

The given vectors are $\text{u}=\langle 8,6\rangle$ and $\text{v}=5\text{i}-10\text{j}$ .

Calculation:

The given vectors can be written as $\text{u}=8\text{i}+6\text{j}$ and $\text{v}=5\text{i}-10\text{j}$ .

Adding the vectors $\text{u}$ and $\text{v}$ .

  $\begin{array}{c}\text{u}+\text{v}=\left(8\text{i}+6\text{j}\right)+\left(5\text{i}-10\text{j}\right)\\ =13\text{i}-4\text{j}\end{array}$

Subtracting the vectors $\text{v}$ from $\text{2u}$ .

  $\begin{array}{c}\text{2u}-\text{v}=2\left(8\text{i}+6\text{j}\right)-\left(5\text{i}-10\text{j}\right)\\ =11\text{i}+22\text{j}\end{array}$

The angle $\theta$ between two nonzero vectors $\text{u}$ and $\text{v}$ .

  $\cos \theta =\frac{\text{u}\cdot \text{v}}{\left|\text{u}\right|\left|\text{v}\right|}$

The magnitude of the vector $\left|\text{w}\right|=\left(a,b\right)$ is given by $\left|\text{w}\right|=\sqrt{a^2+b^2}$ .

Calculate the magnitude of the vector $\text{u}$ .

  $\begin{array}{c}\left|\text{u}\right|=\sqrt{{\left(8\right)}^2+{\left(6\right)}^2}\\ =\sqrt{100}\\ =10\end{array}$

Calculate the magnitude of the vector $\text{v}$ .

  $\begin{array}{c}\left|\text{v}\right|=\sqrt{{\left(5\right)}^2+{\left(-10\right)}^2}\\ =\sqrt{125}\\ =5\sqrt{5}\end{array}$

The dot product of two vectors $\text{u}=\langle 8,6\rangle$ and $\text{v}=\langle 5,-10\rangle$ .

  $\begin{array}{c}\text{u}\cdot \text{v}=\langle 8,6\rangle \cdot \langle 5,-10\rangle \\ =\left(8\right)\left(5\right)+\left(6\right)\left(-10\right)\\ =40-60\\ =-20\end{array}$

Substitute the value in equation $\cos \theta =\frac{\text{u}\cdot \text{v}}{\left|\text{u}\right|\left|\text{v}\right|}$ .

  $\begin{array}{c}\cos \theta =\frac{-20}{\left(10\right)\left(5\sqrt{5}\right)}\\ \cos \theta =\frac{-2}{5\left(5\right)}\\ \theta ={\cos }^{-1}\left(\frac{-2}{5\left(5\right)}\right)\\ =100.30°\end{array}$

The projection of the vectors $\text{u}$ and $\text{v}$ is calculated as.

  ${\text{proj}}_{\text{v}}\text{u}=\left(\frac{\text{u}\cdot \text{v}}{{\left|\text{v}\right|}^2}\right)\text{v}$

Substitute the values in above equation.

  $\begin{array}{c}{\text{proj}}_{\text{v}}\text{u}=\left(\frac{-20}{125}\right)\langle 5,-10\rangle \\ =-\frac{4}{5}\langle 5,-10\rangle \\ =\langle -\frac{20}{25},\frac{40}{25}\rangle \\ =\langle -\frac{4}{5},\frac{8}{5}\rangle \end{array}$

Therefore, the value of is $13\text{i}-4\text{j}$ , $2\text{u}-\text{v}$ is $11\text{i}+22\text{j}$ , the angle between $\text{u}$ and $\text{v}$ is $100.30°$ , and ${\text{proj}}_{\text{v}}\text{u}$ is $\langle -\frac{4}{5},\frac{8}{5}\rangle$ .

(c)

To determine

To find:The work done by $\text{u}$ when a particle moves under its influence from $\left(2,0\right)$ to $\left(10,3\right)$ .

Answer to Problem 5CRT

The work done by $\text{u}$ when a particle moves under its influence from $\left(2,0\right)$ to $\left(10,3\right)$ is $82$ .

Explanation of Solution

Given information:

The given vectors are $\text{u}=\langle 8,6\rangle$ and $\text{v}=5\text{i}-10\text{j}$ .

The particle moves the displacement from $\left(2,0\right)$ to $\left(10,3\right)$ .

Calculation:

The force vector $\text{u}$ is $\text{u}=\langle 8,6\rangle$ .

Suppose particle moving object from $P$ and $Q$ , if the vector $\text{u}$ and $P$ , $Q$ are given as.

  $\text{u}=8\text{i}+6\text{j,}P\left(2,0\right),Q\left(10,3\right)$

The work $W$ done by a force $\text{u}$ in moving along a displacement vector $\text{D}$ is.

  $W=\text{u}\cdot \text{D}$

The dot product of the vector $\text{u}=\langle a_1,b_1\rangle$ and $\text{D}=\langle a_2,b_2\rangle$ is calculated as.

  $\text{u}\cdot \text{D}=a_1{a}_2+b_1{b}_2$

Calculate the displacement between two points $P\left(2,0\right),Q\left(10,3\right)$ .

  $\begin{array}{c}\text{D}=\langle x_2-x_1,y_2-y_1\rangle \\ =\langle \left(10-2\right),\left(3-0\right)\rangle \\ =\langle 8,3\rangle \end{array}$

Calculate the work done.

  $\begin{array}{c}W=\text{u}\cdot \text{D}\\ =\langle 8,3\rangle \cdot \langle 8,6\rangle \\ =\left(8\right)\left(8\right)+\left(3\right)\left(6\right)\\ =82\end{array}$

Therefore, the work done by $\text{u}$ when a particle moves under its influence from $\left(2,0\right)$ to $\left(10,3\right)$ is $82$ .