Use the cross product to find the sin of an angle then prove it

shemer77

1. The problem statement, all variables and given/known data
Leta theta be the angle between the vectors u=2i +3j -6k and v=2i + 3j+6k
A) use the dot product to find cos theta
b) use the cross product to find sin theta
c) confirm that sin^2(theta) + cos^2(theta)=1


3. The attempt at a solution
I got a to be 117 degrees, but however b and c are stumping me, whenever I try to do b I get an the cross product and then try to solve for theta but i get an error using my calculator, I rechecked my calculations twice and everything make sense am I doing something wrong or is it really not possible.

dynamicsolo

It would be more helpful to someone assisting you if you explained what you did for part (b), in order to understand why you are getting an error...

(And I get an angle of 118º for theta.)

shemer77

well all i did was the cross product of u and v which came out to 36i -24j+0k and i took the norm of that which came out to 12*sqrt(13) and I divided that by the norm of u * the norm of v which is 49. so I had 12*sqrt(13)/49 and i took the inverse sin of that. which would obviously give me an error. But I dont see what I did wrong all my calculations are correct.

dynamicsolo

Check how you have input [itex] \frac{12\sqrt{13}}{49} [/itex] ; this is a positive number smaller than 1 . Is your "error" that you don't get the same angle as you did for the arccosine? Don't forget that the range of arccosine is 0º to 180º , but the range of arcsine is -90º to 90º ; the arcsine result for theta (out of a calculator) is the supplement of the correct value. (The angle theta is in the second quadrant.)

Part (c) sidesteps this anyway, since you are simply asked to sum the squares of your values from parts (a) and (b)... [EDIT: You must have gotten cos(theta) = -23/49 to get the angle of 118º . The sum of the squares does give you 1 .]

shemer77

ah awesome man, I got it thanks for your help!!!