# Math Help - How can I find the angle between two vectors?

1. ## How can I find the angle between two vectors?

Hi, can someone help with finding an angle between two vectors using radians?

Here is the problem:

2. Remember,
$||\bold{u}\cdot \bold{v}|| = ||\bold{u}||\cdot ||\bold{v}||\cos \theta$

Use this to find your angle.

3. Yes, I’ve used that method to find the angles and I’ve been successful finding angles when the questions were given as vectors or some multiple of unit vectors. This question happens to be in the form of radians and I haven’t had any luck working it out. I want the answer to be in radian also.

4. Hello, coopsterdude!

Are you sure those i's aren't misplaced?

Find the angle $\theta$ between vectors: . $\begin{array}{ccc}\vec u & = & \cos\frac{\pi}{3} + i\sin\frac{\pi}{3} \\ \vec v & = & \cos\frac{3\pi}{4} + i\sin\frac{3\pi}{4}\end{array}$
Can't you do the trig?

. . $\vec u \; = \; \frac{1}{2} + i\frac{\sqrt{3}}{2} \; = \; \left\langle\frac{1}{2},\,\frac{\sqrt{3}}{2}\right \rangle$

. . $\vec v \; = \; \text{-}\frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2} \; = \;\left\langle \text{-}\frac{\sqrt{2}}{2},\,\frac{\sqrt{2}}{2}\right\ran gle$

Now use that formula . . .

5. Yes, I'm terrible sorry, I did misplace the i's and j's. Thanks for your help, I"ll get working on the problem right away.

6. Hi I tried to solve the problem using the formula but did not get the same answer as in my book. So I used the unit circle and visually noted the difference between the two points on the unit circle which was 75. After converting 75 degrees into radian, my answer was 5pi/12 which is the correct answer. I still don’t know why I couldn’t find my answer using any of the formulas in my book or your example.

7. Write the vectors is this form: $u = \frac{1}{2}i + \frac{{\sqrt 3 }}{2}j\quad \& \quad v = - \frac{{\sqrt 2 }}{2}i + \frac{{\sqrt 2 }}{2}k.$
Now it is clear that both are unit vectors.

Also it follows that $\cos (\phi ) = \frac{{u \cdot v}}{{\left\| u \right\|\left\| v \right\|}} = - \frac{{\sqrt 2 }}{4} + \frac{{\sqrt 6 }}{4}$.

But, $\phi = \arccos \left( { - \frac{{\sqrt 2 }}{4} + \frac{{\sqrt 6 }}{4}} \right) = \frac{{5\pi }}{{12}}.$