Chapter 9, Problem 18RE

To determine

To find: Whether the given two vectors are orthogonal or not if not, then find the angle between them.

Answer to Problem 18RE

The angle between the two given vector is $\theta =77.47°$ .

Explanation of Solution

Given information:

The given vectors are:

  $\begin{array}{l}u=\langle \text{5,3}\rangle \\ v=\langle -2\text{,6}\rangle \end{array}$

Concept used:

The two vectors $u=\langle a,b\rangle$ & $v=\langle c,d\rangle$ are orthogonal if,

  $u\cdot v=0$

Angle between two vectors $u=\langle a,b\rangle$ & $v=\langle c,d\rangle$ is given,

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

The dot product of given vectors $u$ & $v$ is calculated as,

  $\begin{array}{c}u\cdot v=5\times \left(-2\right)+3\times 6\\ =-10+18\\ =8\end{array}$

As, $u\cdot v\ne \text{0}$ , so $u$ & $v$ are not orthogonal,

Further simplified as,

Angle between of given vectors $u$ & $v$ is calculated as,

  $\begin{array}{c}\left|u\right|=\sqrt{5^2+3^2}\\ =\sqrt{25+9}\\ =\sqrt{34}.\end{array}$

  $\begin{array}{c}\left|v\right|=\sqrt{-2^2+6^2}\\ =\sqrt{4+36}\\ =\sqrt{40}\end{array}$

  $\begin{array}{c}\cos \theta =\frac{8}{\sqrt{34}\times \sqrt{40}}\\ =\frac{8}{4\sqrt{85}}\\ =\frac{2}{\sqrt{85}}\end{array}$

Further simplified as,

  $\theta ={\cos }^{-1}\left(\frac{2}{\sqrt{85}}\right)$

  $=77.47°$

Therefore, the angle between the two given vector is $\theta =77.47°$ .