Chapter 6.2, Problem 38E

Solution

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

Parallel

Given:

  $\overrightarrow{u}=〈2,-7〉,\overrightarrow{v}=〈-4,14〉$

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:

  $\begin{array}{l}\overrightarrow{u}.\overrightarrow{v}=〈2,-7〉.〈-4,14〉\\ \overrightarrow{u}.\overrightarrow{v}=\left(2\right)\left(-4\right)+\left(-7\right)\left(14\right)\\ \overrightarrow{u}.\overrightarrow{v}=-106\end{array}$

The magnitudes are:

  $\begin{array}{l}\left|\overrightarrow{u}\right|=\sqrt{{\left(2\right)}^2+{\left(-7\right)}^2}=\sqrt{53}\\ \left|\overrightarrow{v}\right|=\sqrt{{\left(-4\right)}^2+{\left(14\right)}^2}=2\sqrt{53}\end{array}$

The calculation for the angle:

  $\begin{array}{l}\theta ={\cos }^{-1}\left(\frac{\overrightarrow{u}.\overrightarrow{v}}{\left|\overrightarrow{u}\right|\left|\overrightarrow{v}\right|}\right)\\ \theta ={\cos }^{-1}\left(\frac{-106}{\left(\sqrt{53}\right)\left(2\sqrt{53}\right)}\right)\\ \theta \approx {180}^{°}\end{array}$

Hence, parallel