Finding the Angle Between Two Vectors

[formulajoe]

ive got two vectors and i need to find the angle between them.
AB = -1.95i + 2.4j +.3k
AC = 2.4j+1.8k
i looked up the answer and found it to be 46.8 deg.
i used the equation (AB)(AC) cos theta = ACxABx +ACyABy + ACzABz
the dot product is 6.84. but the equation ends up like this
cos theta = (ACxABx +ACyABy + ACzABz)/(AB)(AC)

this is a dot product on top and on bottom right? i don't quite understand how this works. according to the answer, ABAC should be around 10.

 

[gnome]

I get a slightly different answer. (I hope I haven't forgotten everything from last semester.)

AC_x is 0 so the dot product is 0 + (2.4)(2.4) + (.3)(1.8) = 6.3

and, the denominator of your fraction is not a dot product. It is the product of the MAGNITUDES of the two vectors.

Do you know how to get those?

Based on this I got θ = 47.5^o (approx)

 

[NateTG]

If you have
$$\vec{a}=<a_1,a_2,a_3>$$
and
$$\vec{b}=<b_1,b_2,b_3>$$

then the angle is
$$\cos^{-1}\frac{a_1b_1+a_2b_2+a_3b_3}{ \sqrt{a_1^2+a_2^2+a_3^2} \sqrt{b_1^2+b_2^2+b_3^2}}$$

 

[NanoTech]

argh, wish i had that this morning. we never went over that, he always gave us the angle already.