# vector angle

• May 10th 2009, 10:58 AM
djmccabie
vector angle
two forces $F_1$N and $F_2$ N are given by

$F_1 = 24i + 32j - 42k$

$F_2 = -3i - 72j - 30k$

a) calculate, correct to the nearest degree, the angle between the directions of $F_1$ and $F_2$

so using the dot product formula $F_1 \cdot F_2 = |F_1||F_2|cos\theta$

I end up with $\frac{-1332}{58 * 3\surd677} = cos\theta$

$\therefore \theta = 72.89$

I think i have done this right however the angle i inverse cos is negative, so do I draw up a cast diagram to find the angle? if so is it the angle 180 - 72.98?
• May 10th 2009, 11:10 AM
Plato
I think that you made a mistake in the dot product. I get $-1116$.

Recall that if the dot product is negative the angle is obtuse.
• May 10th 2009, 11:22 AM
djmccabie
yes i mis-calculated. i now get the angle 75.7.

so I'm thinking I do 180-75.7 = 104.3 ?