Finding the angle in between 3D vectors

[pugfug90]

Homework Statement

I did the rest of a problem right.. It asks to determine whether the parallelogram is a rectangle, meaning right angles..

A(2,-1,4), B(3,1,2), C(0,5,6), D(-1,3,8)

Homework Equations

cos theta = dot product of u and v divided by magnitude of u and v

The Attempt at a Solution

I drew a sketch parallelogram, with A at Northwest, B at NE, C at SE, and D at SW..
blah blah
I got vector AD and BC to be <-3,4,4> and vectors AB and DC to be <1,2,-2>

mag of AB/BC is square root 41, etc. if you can help me, you should know how I got there.
now..

cos theta = [(-3*1) +(4*2) + (4*-2)]/[square root of 41*square root of 9]

[-3]/[square root 369]=cos theta

theta=?98.9?

The back of my book says 81.02 degrees..

 

[christianjb]

You're not doing anything wrong. There are two different angles in a parallelogram which sum to 180 degrees. 98.9+81.1= 180.

 

[pugfug90]

Thanks for that! :-D