# Vector \vec{A} is 3.00 units in length and points along

Vector $\stackrel{\to }{A}$ is 3.00 units in length and points along the positive x-axis. Vector $\stackrel{\to }{B}$ is 4.00 units in tength and points along the negative y-axis. Use graphical methods to find the magnitude and direction of the vectors $\stackrel{\to }{A}+\stackrel{\to }{B}$
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

GaceCoect5v
Since we are asked to use a graphical method to solve the problem, we need to choose an appropriate scale to plot the vectors then measure the resulting vector from the addition of $\stackrel{\to }{A}$ and $\stackrel{\to }{B}$. In the graph below, each square has a side of a length of 1 unite.

We notice that the length of the vector $\left(\stackrel{\to }{A}+\stackrel{\to }{B}\right)$ in the graph below is about 5 squares, meaning that its about 5 unites in length. The angle between $\left(\stackrel{\to }{A}+\stackrel{\to }{B}\right)$ and the positive x-axis can be calculated as follows:
$\theta ={\mathrm{tan}}^{-1}\left(\frac{midBmid}{midAmid}\right)={\mathrm{tan}}^{-1}\left(\frac{4}{3}\right)=53.13°$
However, notice that $\left(\stackrel{\to }{A}+\stackrel{\to }{B}\right)$ is in the fourth quarter, which means that $\theta =-53.13°$.
$mid\stackrel{\to }{A}+\stackrel{\to }{B}mid=5$ unites and makes an angle $\theta =-53.13°$ with the positive x-axis

Toni Scott
Explanation:
Vectors in two dimensions have two components Y and X, and are written in the following form:
(X,Y)
In this case we have two vectors:
$\stackrel{\to }{A}=\left(3,0\right)$
$\stackrel{\to }{B}=\left(0,-4\right)$
We need to find $\stackrel{\to }{A}+\stackrel{\to }{B}$ and $\stackrel{\to }{A}-\stackrel{\to }{B}$, their magnitude and direction.
For the magnitude we will use the formula to calculate the distance d between two points (pithagorean theorem):
$d=\sqrt{{\left(Y2-Y1\right)}^{2}+{\left(X2-X1\right)}^{2}}$ (1)
For the direction, which is the measure of the angle the vector makes with a horizontal line, we will use the following formula:
$\mathrm{tan}\theta =\frac{Y2-Y1}{X2-X1}$ (2)
Now, lets