Chapter 2.9, Problem 8PP

Do not calculate the result.

Prob. P2-8

To determine

To set up:

The dot product to calculate the angle $\theta$.

Answer to Problem 8PP

The setup of the dot product in case (a) is $\underset{\_}{-3=3\left(3\right)\cos \theta }$.

Explanation of Solution

Establish the coordinates for point $O\left(0,0,0\right)\text{m}$, $A\left(0,0,3\right)\text{m}$, and point $B\left(2,2,-1\right)\text{m}$.

Express the position vector r in the direction of OA.

$r_{OA}=\left(x_A-x_O\right)i+\left(y_A-y_O\right)j+\left(z_A-z_O\right)k$ (I)

Here, the coordinates of A and O are $x_A,x_O,y_A,y_O,z_A,\text{and}{z}_O$, respectively.

Express the magnitude of the direction of OA.

$r_{OA}=\sqrt{{\left(x_A-x_O\right)}^2+{\left(y_A-y_O\right)}^2+{\left(z_A-z_O\right)}^2}$ (II)

Express the position vector r in the direction of OB.

$r_{OB}=\left(x_B-x_O\right)i+\left(y_B-y_O\right)j+\left(z_B-z_O\right)k$ (III)

Here, the coordinates of B and O are $x_B,x_O,y_B,y_O,z_B,\text{and}{z}_O$, respectively.

Express the magnitude of the direction of OB.

$r_{OB}=\sqrt{{\left(x_B-x_O\right)}^2+{\left(y_B-y_O\right)}^2+{\left(z_B-z_O\right)}^2}$ (IV)

Determine the angle between two vectors.

$\cos \theta =\frac{r_{OA}\cdot r_{OB}}{r_{OA}\cdot r_{OB}}$ (V)

Conclusion:

Substitute 0 for $x_A$, 0 for $y_A$, 3 m for $z_A$, 0 m for $x_O$, 0 m for $y_O$, and 0 m for $z_O$ in Equation (I).

$\begin{array}{c}{r}_{OA}=\left(0-0\right)i+\left(0-0\right)j+\left(3-0\right)k\\ =\left\{3k\right\}\text{m}\end{array}$

Substitute 0 for $x_A$, 0 for $y_A$, 3 m for $z_A$, 0 m for $x_O$, 0 m for $y_O$, and 0 m for $z_O$ in Equation (II).

$\begin{array}{c}{r}_{OA}=\sqrt{{\left(0-0\right)}^2+{\left(0-0\right)}^2+{\left(3-0\right)}^2}\\ =\sqrt{9}\\ =3\text{m}\end{array}$

Substitute 2 m for $x_B$, 2 m for $y_B$, ­1 m for $z_B$, 0 m for $x_O$, 0 m for $y_O$, and 0 m for $z_O$ in Equation (III).

$\begin{array}{c}{r}_{OB}=\left(2-0\right)i+\left(2-0\right)j+\left(-1-0\right)k\\ =\left\{2i+2j-1k\right\}\text{m}\end{array}$

Substitute 2 m for $x_B$, 2 m for $y_B$, ­1 m for $z_B$, 0 m for $x_O$, 0 m for $y_O$, and 0 m for $z_O$ in Equation (IV).

$\begin{array}{c}{r}_{OB}=\sqrt{{\left(2-0\right)}^2+{\left(2-0\right)}^2+{\left(-1-0\right)}^2}\\ =\sqrt{4+4+1}\\ =3\text{m}\end{array}$

Substitute $\left\{3k\right\}\text{m}$ for $r_{OA}$, $\left\{2i+2j-1k\right\}\text{m}$ for $r_{OB}$, 3 m for $r_{OA}$, and 3 m for $r_{OB}$ in Equation (V).

$\begin{array}{l}\cos \theta =\frac{\left\{3k\right\}\text{m}\cdot \left\{2i+2j-1k\right\}\text{m}}{3\text{m}\cdot 3\text{m}}\\ -3=3\left(3\right)\cos \theta \end{array}$

Thus, the set-up of the dot product in case (a) is $\underset{\_}{-3=3\left(3\right)\cos \theta }$.

To determine

To set up:

The dot product to calculate the angle $\theta$

Answer to Problem 8PP

The setup of the dot product in case (b) is $\underset{\_}{-5=3\left(2.5\right)\cos \theta }$.

Explanation of Solution

Let us establish the coordinates for point $O\left(0,0,0\right)\text{m}$, $A\left(-2,2,1\right)\text{m}$, and point $B\left(1.5,0,-2\right)\text{m}$.

Conclusion:

Substitute ­2 m for $x_A$, 2 m for $y_A$, 1 m for $z_A$, 0 m for $x_O$, 0 m for $y_O$, and 0 m for $z_O$ in Equation (I).

$\begin{array}{c}{r}_{OA}=\left(-2-0\right)i+\left(2-0\right)j+\left(1-0\right)k\\ =\left\{-2i+2j+k\right\}\text{m}\end{array}$

Substitute ­2 m for $x_A$, 2 m for $y_A$, 1 m for $z_A$, 0 m for $x_O$, 0 m for $y_O$, and 0 m for $z_O$ in Equation (II).

$\begin{array}{c}{r}_{OA}=\sqrt{{\left(-2-0\right)}^2+{\left(2-0\right)}^2+{\left(1-0\right)}^2}\\ =\sqrt{4+4+1}\\ =3\text{m}\end{array}$

Substitute 1.5 m for $x_B$, 0 m for $y_B$, ­2 m for $z_B$, 0 m for $x_O$, 0 m for $y_O$, and 0 m for $z_O$ in Equation (III).

$\begin{array}{c}{r}_{OB}=\left(1.5-0\right)i+\left(0-0\right)j+\left(-2-0\right)k\\ =\left\{1.5i-2k\right\}\text{m}\end{array}$

Substitute 1.5 m for $x_B$, 0 m for $y_B$, ­2 m for $z_B$, 0 m for $x_O$, 0 m for $y_O$, and 0 m for $z_O$ in Equation (IV).

$\begin{array}{c}{r}_{OB}=\sqrt{{\left(1.5-0\right)}^2+{\left(0-0\right)}^2+{\left(-2-0\right)}^2}\\ =\sqrt{2.25+0+4}\\ =2.5\text{m}\end{array}$

Substitute $\left\{-2i+2j+k\right\}\text{m}$ for $r_{OA}$, $\left\{1.5i-2k\right\}\text{m}$ for $r_{OB}$, 3 m for $r_{OA}$, and 2.5 m for $r_{OB}$ in Equation (V).

$\begin{array}{l}\cos \theta =\frac{\left\{-2i+2j+k\right\}\text{m}\cdot \left\{1.5i-2k\right\}\text{m}}{3\text{m}\cdot 2.5\text{m}}\\ -5=3\left(2.5\right)\cos \theta \end{array}$

Thus, the set-up of the dot product in case (b) is $\underset{\_}{-5=3\left(2.5\right)\cos \theta }$.