Question

Let IuI=2 and IvI=3, and let the angle between the two vectors be 2pi/3.
a) Find u.v
b)Find I2u-3vI

Answer 1

\[
u\cdot v = |u||v|\cos \theta
\]

Answer 2

\(\bf {\color{blue}{ |u|=2\quad |v|=3\qquad \theta = \cfrac{2\pi}{3}\qquad
\begin{array}{llll}
\textit{angle between two vectors }\\ \quad \\
cos(\theta)=\cfrac{u \cdot v}{||u||\ ||v||}
\end{array}}}\\ \quad \\

cos(\theta)=\cfrac{u \cdot v}{||u||\ ||v||}\implies cos(\theta)\cdot ||u||\ ||v||=u \cdot v \)

Answer 3

What about I2u-3vI ?

Answer 4

well.. didn't get far on that one... maybe wio...
I was wondering how to solve for either, then you'd have the other to deal with

Answer 5

Well, there are two angles here. If \(u\) has angle \(\phi\), then \(v\) has angle \(\phi \pm \theta\). The question seems to imply that \(\phi\) does not matter. So in that case, just assume that \(\phi=0\).

Answer 6

Then you can say \[
\mathbf u = \langle 2,0 \rangle
\]And\[
\mathbf v = \langle 3\cos(\theta), 3\sin(\theta) \rangle
\]

Answer 7

Then you can directly compute it.

Answer 8

It's a bit of a dirty method, because you are making an assumption about \(\phi\) by leveraging the facts about the question. Despite that, it will work.

Alternatively, you can let \(\phi = x\) and try to solve in terms of \(x\) and rigorously show that the answer is independent of \(x\) and that would be more mathematically sound.

Or, you could try to show pictorially that rotating \(u\) and \(v\) will only rotate \(2u-3v\), and not change its magnitude, thus it is legitimate to rotate \(u\) to be on the \(x\) axis.

Answer 9

@jdoe0001 @JonC Does this help?

Answer 10

well.... a bit ambiguous maybe =), but helps

Answer 11

Ambiguous? What is unclear?