Chapter 6.2, Problem 35E

Solution

To Find: whether the vectors $\overrightarrow{u}$ and $\overrightarrow{v}$ are parallel, orthogonal, or neither.

The vectors are neither parallel nor orthogonal each other.

Given:

  $\overrightarrow{u}=〈15,-12〉,\overrightarrow{v}=〈-4,5〉$

Concept Used:

When the dot product of two vector is zero then the vectors are orthogonal.

If the vectors are not orthogonal then the angle tells whether the pair of vectors are parallel $\left(0^{°}and{180}^{°}\right)$ or neither $\left(\theta \ne 0^{°}or{180}^{°}\right)$ .

Calculation:

  $\begin{array}{l}\overrightarrow{u}.\overrightarrow{v}=〈15,-12〉.〈-4,5〉\\ \overrightarrow{u}.\overrightarrow{v}=\left(15\right)\left(-4\right)+\left(-12\right)\left(5\right)\\ \overrightarrow{u}.\overrightarrow{v}=-120\end{array}$

The magnitudes are:

  $\begin{array}{l}\left|\overrightarrow{u}\right|=\sqrt{{\left(15\right)}^2+{\left(-12\right)}^2}=3\sqrt{41}\\ \left|\overrightarrow{v}\right|=\sqrt{{\left(-4\right)}^2+{\left(5\right)}^2}=\sqrt{41}\end{array}$

The calculation for the angle:

  $\begin{array}{l}\theta ={\cos }^{-1}\left(\frac{\overrightarrow{u}.\overrightarrow{v}}{\left|\overrightarrow{u}\right|\left|\overrightarrow{v}\right|}\right)\\ \theta ={\cos }^{-1}\left(\frac{-120}{\left(3\sqrt{41}\right)\left(\sqrt{41}\right)}\right)\\ \theta \approx {167.32}^{°}\end{array}$

Hence, the vectors are neither parallel nor orthogonal each other.