Chapter 1, Problem 1.64P

Determine (a) the angle between the position vectors P and Q; and (b) a unit vector perpendicular to P and Q.

To determine

(a)

The angle between the position vectors P and Q.

Answer to Problem 1.64P

The angle between the position vectors P and Q is, $\theta ={44.312}^0.$

Explanation of Solution

Given Information:

Consider the given diagram

  $\begin{array}{l}\overrightarrow{P}=\left(3-0\right)\overline{i}+\left(2-2\right)\overline{j}\left(4-0\right)\overline{k}\\ \overrightarrow{P}=3\overline{i}+4\overline{k}\end{array}$

Similarly Q is

  $\begin{array}{l}\overrightarrow{Q}=\left(0-0\right)\overline{i}+\left(0-2\right)\overline{j}+\left(4-0\right)\overline{k}\\ \overrightarrow{Q}=-2\overline{j}+4\overline{k}\end{array}$

Angle between the vectors is

  $\begin{array}{l}\cos \theta =\frac{\overrightarrow{P}.\overrightarrow{Q}}{\left|\overline{P}\right|\left|\overline{Q}\right|}\\ \cos \theta =\frac{\left(3\overline{i}+4\overline{k}\right).\left(-2\overline{j}+\overline{4k}\right)}{\sqrt{{\left(3\right)}^2+{\left(4\right)}^2}.\sqrt{{\left(-2\right)}^2+\left(4\right)}}\\ \cos \theta =\frac{16}{5\sqrt{20}}\\ \cos \theta =0.7155\\ \theta ={44.312}^{°}\end{array}$

To determine

(b)

The unit vector perpendicular to P and Q.

Answer to Problem 1.64P

The unit vector perpendicular to P and Q is, $0.512\overline{i}-0.768\overline{j}-0.384\overline{k}.$

Explanation of Solution

Given Information:

Consider the given diagram

The cross product of the two vectors given vector perpendicular to P and Q

  $\begin{array}{l}\overrightarrow{C}=\overrightarrow{a}\times \overrightarrow{b}\\ \overrightarrow{C}=\overline{P}\times \overline{Q}\\ \overrightarrow{C}=\left(\overline{3i}+\overline{4k}\right)\times \left(\overline{-2j}+\overline{4k}\right)\\ =\left|\overline{\begin{array}{l}i\\ 3\\ 0\end{array}}\overline{\begin{array}{l}j\\ 0\\ -2\end{array}}\overline{\begin{array}{l}k\\ 4\\ 4\end{array}}\right|\\ =\overline{i}\left(8\right)-\overline{j}\left(12\right)+\overline{k}\left(6\right)\\ =8\overline{i}-12\overline{j}+6\overline{k}\end{array}$

Then the unit vector is

  $\begin{array}{l}=\frac{8\overline{i}-12\overline{j}+6\overline{k}}{\sqrt{8^2+{12}^2+6^2}}\\ =\frac{1}{\sqrt{244}}\left(8\overline{i}-12\overline{j}+6\overline{k}\right)\\ =\frac{\left(8\overline{i}-12\overline{j}+6\overline{k}\right)}{15.62}\\ =0.512\overline{i}-0.768\overline{j}-0.384\overline{k}\end{array}$