Chapter 20, Problem 1RE

To determine

The magnitude and direction of the vector in standard position with its end points at $\left(4,6\right)$.

Answer to Problem 1RE

The magnitude and direction of the vector in standard position with end points at $\left(4,6\right)$ is $\underset{\_}{7.211\text{ and }56.3°}$ respectively.

Explanation of Solution

Result used:

“The magnitude in standard position for a vector v with end points at $\left(x,y\right)$ is, ${\Vert v\Vert }^2=x^2+y^2$.”

“The direction of the vector in standard position with given end points is calculated by substituting the coordinates of the end points $\left(x,y\right)$ into tangent function, $\tan \theta =\frac{y}{x}\text{ or }\theta ={\tan }^{-1}\left(\frac{y}{x}\right)$.”

Calculation:

Given that, the coordinates of the vector is $\left(4,6\right)$.

Substitute $x=4$ and $y=6$ in the formula ${\Vert v\Vert }^2=x^2+y^2$ and obtain the magnitude of the vector as follows.

$\begin{array}{c}{\Vert v\Vert }^2=4^2+6^2\\ =16+36\\ =52\\ \Vert v\Vert =\pm \sqrt{52}\\ =\pm 7.211\end{array}$

As the magnitude is positive, so the magnitude of the vector is 7.211.

Therefore, the magnitude of the vector in standard position with end points at $\left(4,6\right)$ is $\underset{\_}{7.211}$.

Substitute $x=4$ and $y=6$ in $\tan \theta =\frac{y}{x}$ and solve for $\theta$ to obtain the direction of the vector as follows.

$\begin{array}{l}\tan \theta =\frac{6}{4}\\ \theta ={\tan }^{-1}\left(\frac{6}{4}\right)\\ \theta ={\tan }^{-1}\left(\frac{3}{2}\right)\\ \theta =56.3°\end{array}$

Thus, the direction of the vector in standard position with end points at $\left(4,6\right)$ is $\underset{\_}{56.3°}$.