# Vectors and Magnitudes

• Feb 15th 2009, 12:29 PM
92stealth
Vectors and Magnitudes
Find the component form of a vector v given the magnitude of u and u+v and the angles that u and u+v make with the positive x-axis.

sorry, fixed.....
||u||=2, $\displaystyle \theta$ = 230, ||u+v|| = 8 $\displaystyle \theta$ = 80

I can't find any problem like this in the book, instructor gave us this as a take home problem to work on.... (Thinking)

• Feb 15th 2009, 12:36 PM
Legendsn3verdie
red x bro.. can't see anything..
• Feb 15th 2009, 12:40 PM
Reckoner
Quote:

Originally Posted by 92stealth
Find the component form of a vector v given the magnitude of u and u+v and the angles that u and u+v make with the positive x-axis.

If you have the magnitude of $\displaystyle \textbf u$ and $\displaystyle \textbf u+\textbf v,$ along with their angles, it should be easy to get their component forms. Once you have that, $\displaystyle \textbf v$ is simply $\displaystyle \textbf (\textbf u+\textbf v)-\textbf u\text.$

You could also set it up as a triangle with sides of length $\displaystyle \lVert\textbf u\rVert, \lVert\textbf v\rVert,$ and $\displaystyle \lVert\textbf u+\textbf v\rVert,$ and angle $\displaystyle |\theta_1-\theta_0|.$ You can use the law of cosines to get $\displaystyle \lVert\textbf v\rVert\text.$

Edit: Now that you have posted the actual values, I see that your problem is much simpler. $\displaystyle \textbf u$ and $\displaystyle \textbf u+\textbf v,$ make the same angle with the $\displaystyle x$-axis, which means that they are parallel. What does that tell you about $\displaystyle \textbf v?$
• Feb 15th 2009, 12:52 PM
92stealth
wow, i am 100% not on my game today. I messed up that first post once again, should have been 80 degrees on the second -.- sorry