0
$\begingroup$

enter image description here

So far I have figured out everything except the angle alpha of f,g. What I tried was I drew this triangle:

enter image description here

and found side length c by doing dist(f,g) = ||f-g|| = sqrt() and I ended up with c = sqrt(54)

Then I used the law of cosines to find the angle of alpha and got 14.14 degrees but it was the wrong answer.

What is wrong with my approach?

  • 0
    $\frac{-11}{4\sqrt{39}}$ was the correct answer, and yes it is.2012-04-09

1 Answers 1

1

I believe the angle is arccos(| < f, g > | / (||f|| * ||g||)) So assuming your numbers are correct that would be 1.115r or ~64deg

  • 1
    I'm not allowed to comment in the above thread, but you could try it with the negative. That yields 116deg. I thought it was abs on the top value though.2012-04-09