Question

question
the vector $overrightarrow{ab}$ has initial point $a(5, - 1)$ and terminal point $b(-4,8)$. find the magnitude and direction of $overrightarrow{ab}$. enter the magnitude in simplest radical form and angle $\theta$ in degrees.
note: do not forget to add a degree symbol for the $\theta$.
provide your answer below:
$|overrightarrow{ab}|=square;\theta=square$

Explanation:

Step1: Find the components of the vector

If $A(5,-1)$ and $B(-4,8)$, then $\overrightarrow{AB}=\langle -4 - 5,8-(-1)
angle=\langle - 9,9
angle$.

Step2: Calculate the magnitude of the vector

The magnitude of a vector $\overrightarrow{v}=\langle x,y
angle$ is $|\overrightarrow{v}|=\sqrt{x^{2}+y^{2}}$. For $\overrightarrow{AB}=\langle -9,9
angle$, $|\overrightarrow{AB}|=\sqrt{(-9)^{2}+9^{2}}=\sqrt{81 + 81}=\sqrt{162}$.

Step3: Calculate the direction of the vector

The direction $\theta$ of a vector $\overrightarrow{v}=\langle x,y
angle$ is given by $\tan\theta=\frac{y}{x}$. Here, $x=-9$ and $y = 9$, so $\tan\theta=\frac{9}{-9}=-1$. Since the vector $\overrightarrow{AB}$ has a negative $x$ - component and a positive $y$ - component, it lies in the second quadrant. So $\theta = 135^{\circ}$.

Answer:

$|\overrightarrow{AB}| = \sqrt{162}$; $\theta = 135^{\circ}$