# Linear Algebra - Inner Product Spaces Question

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

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?

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I'm still getting $\sqrt{54}$, $\int_{0}^{1}((5x^2-3)-(6x+3))*((5x^2-3)-(6x+3)) dx = 54$ then square root is $\sqrt{54}$... – StickFigs Apr 9 '12 at 2:06
You’re right: I miscopied one of the functions. The problem is in your calculation of $\alpha$. I get $\cos\alpha=\frac{11}{4\sqrt{39}}\approx 0.44$, $\alpha\approx 1.1148$ radians (or about 63.87°). – Brian M. Scott Apr 9 '12 at 2:14
Those answers are incorrect, see my other comment below. – StickFigs Apr 9 '12 at 2:55
I apologize: I seem to be brain-dead tonight. I dropped a minus sign: $\cos\alpha=\frac{-11/2}{2\sqrt{39}}=-\frac{11}{4\sqrt{39}}$. By the way, I’d expect it to want the angle in radians: that’s a nearly universal convention in mathematics. Out of curiosity, is this WeBWorK? – Brian M. Scott Apr 9 '12 at 3:11
$\frac{-11}{4\sqrt{39}}$ was the correct answer, and yes it is. – StickFigs Apr 9 '12 at 15:46