What is the correct angle between T and U in the cross product?

[Linus Pauling]

1. Find the sine of the angle between and .

2. AxB = ABsin(alpha)



3. I know T = (3,1,0) and U = (2,4,0), and I calculated that V = TxU = (0,0,10), so the magnitude of V is 10.

Then I did 10 = sqrt(10)*sqrt(20)*sin(alpha)

Divided 10 by sqrt(10)*sqrt(20), took the inverse sine, and obtained 45 degrees, which is incorrect.

 

[Mentallic]

I haven't studied 3-D vectors, but this problem can be simplified into a 2-D diagram, and quite easily since z=0 for both vectors.
I also got the answer of 45^o by answering it with trigonometry.

But the question does say find the sine of the angle, correct? So are we looking for the answer 1/\sqrt{2} ?

 

[Linus Pauling]

We are not finding sine(theta), we are finding theta, which is what I calculated.

 

[HallsofIvy]

You have already been told that 45 degrees (\pi/4 radians) is the correct answer. What makes you say it is incorrect?

As a check, use the fact that \vec{u}\cdot\vec{v}= |\vec{u}||\vec{v}|cos(\theta). Since, here, |\vec{u}\times\vec{v}|= \vec{u}\cdot\vec{v}, sin(\theta)= cos(\theta) and the angle, being in the first quadrant must be \pi/4.