# Basic vectors help ...

• Apr 10th 2008, 04:45 AM
struck
Basic vectors help ...
I need to find the resultant vector c in magnitude and direction.

I found the magnitude to be 7.2 using cosine rule, i.e.:

$\displaystyle c^2 = a^2 + b^2 - 2ab cos C$

However, I'm not able to find the magnitude.

So far, I have tried the following:

Since resultant vector c = a + b

ax = -(6 cos 40)
ay = 6

bx = 3 cos 70
by = 3

Then,
cx = ax + bx
cy = ay + by

But that doesn't seem to help. (Thinking)
• Apr 10th 2008, 05:10 AM
earboth
Quote:

Originally Posted by struck
I need to find the resultant vector c in magnitude and direction.

I found the magnitude to be 7.2 using cosine rule, i.e.: Unfortunately wrong

$\displaystyle c^2 = a^2 + b^2 - 2ab cos C$

So far, I have tried the following:

Since resultant vector c = a + b Correct

...

I've modified your drawing a little bit.

I placed the origin at the beginning of vector $\displaystyle \vec a$. Using the indicated angle you'll get:

$\displaystyle \vec a = 6 \cdot \left(\begin{array}{c}\cos(140^\circ) \\ \sin(140^\circ)\end{array} \right)$ ....... and ....... $\displaystyle \vec b = 3 \cdot \left(\begin{array}{c}\cos(70^\circ) \\ \sin(70^\circ)\end{array} \right)$

Now the next steps are quite easy.

I've got $\displaystyle |\vec c| \approx 7.57$
• Apr 10th 2008, 05:30 AM
struck
Edit: errr.. never mind.

I actually got 7.6, but wrote 7.2 in my first post here.. So I was correct at least about that... Thanks