## What is the direction of vector A + vector B? Feb 4, 2018

$- {63.425}^{o}$

#### Explanation: Not drawn to scale

Sorry for the crudely drawn diagram but I hope it helps us see the situation better.

As you have worked out earlier in the question the vector:

$A + B = 2 i - 4 j$

in centimeters. To get the direction from the x-axis we need the angle. If we draw the vector and split it up into its components, i.e. $2.0 i$ and $- 4.0 j$ you see we get a right angled triangle so the angle can be worked out using simple trigonometry. We have the opposite and the adjacent sides. From trigonometry:

$\tan \theta = \frac{O p p}{A \mathrm{dj}} \implies \theta = {\tan}^{-} 1 \left(\frac{O p p}{A \mathrm{dj}}\right)$

In our case the side opposite the angle is $4.0 c m$ so $4.0 c m$ and the adjacent side is: $2.0 c m$ so:

$\theta = {\tan}^{-} 1 \left(\frac{4.0}{2.0}\right) = {63.425}^{o}$

Obviously this is anti-clockwise so we must put a minus in front of the angle $\to - 63.425$

If the question is asking for the positive angle going clockwise around the diagram then simple subtract this from ${360}^{o}$

$\to 360 - 63.425 = {296.565}^{o}$

Feb 4, 2018

e. ${296.5}^{\circ}$
f. ${0}^{\circ}$

#### Explanation:

Note: I am using the angle measuring method in which you start at the +x axis and circulate counterclockwise to the vector. So the +y axis is at ${90}^{\circ}$ and the minus y axis is at ${270}^{\circ}$. Ref:
http://chortle.ccsu.edu/VectorLessons/vch05/vch05_3.html

e. From your work, $\vec{A} + \vec{B} = 2 \text{ cm " hati - 4 " cm } \hat{j}$. That puts the vector in the 4th quadrant. Draw the vector with the arrowhead at x=2, y=-4.

Let's calculate the angle ${\theta}_{e}$ between the -y axis and the vector. The length of the side opposite is 2 cm and the side adjacent is 4 cm.

${\tan}^{-} 1 \left(\frac{2}{4}\right) = {26.5}^{\circ}$

The -y axis is already ${270}^{\circ}$ counterclockwise from the +x axis, so the answer to e is ${270}^{\circ} + {26.5}^{\circ} = {296.5}^{\circ}$.

f. From your work, $\vec{A} - \vec{B} = 4 \text{ cm " hati + 0 " cm } \hat{j}$. Therefore the resultant lies along the x axis. That is an angle of ${0}^{\circ}$.

I hope this helps,
Steve