Question

What are the magnitude and direction of w = ❬–10, –12❭? Round your answer to the thousandths place.

Answer 1

The direction of a vector is the orientation of the vector, that is, the angle it makes with the x-axis.

The magnitude of a vector is its length.

The formulas to find the magnitude and direction of a vector are:

$\begin{gathered} u=❬x,y❭\Rightarrow\text{ Vector} \\ \mleft\Vert u|\mright|=\sqrt[]{x^2+y^2}\Rightarrow\text{ Magnitude} \\ \theta=\tan ^{-1}(\frac{y}{x})\Rightarrow\text{ Direction} \end{gathered}$

In this case, we have:

• Magnitude

$\begin{gathered} w=❬-10,-12❭ \\ \Vert w||=\sqrt[]{(-10)^2+(-12)^2} \\ \Vert w||=\sqrt[]{100+144} \\ \Vert w||=\sqrt[]{244} \\ \Vert w||\approx15.620\Rightarrow\text{ The symbol }\approx\text{ is read 'approximately'} \end{gathered}$

• Direction

$\begin{gathered} w=❬-10,-12❭ \\ \theta=\tan ^{-1}(\frac{-12}{-10}) \\ \theta=\tan ^{-1}(\frac{12}{10}) \\ \theta\approx50.194\text{\degree} \\ \text{ Add 180\degree} \\ \theta\approx50.194\text{\degree}+180\text{\degree} \\ \theta\approx230.194\text{\degree} \end{gathered}$

Therefore, the magnitude and direction of the vector are:

$\begin{gathered} \Vert w||\approx15.620 \\ \theta\approx230.194\text{\degree} \end{gathered}$