Cross Product Magnitude and Direction Calculation for Vector v and Vector B

[togame]

Homework Statement

Find the magnitude of \vec{v} \times \vec{B} and the direction of the new vector.
\vec{v}=\left \langle 0,0,20 \right \rangle
\vec{B}=\left \langle 1,1,0 \right \rangle

Homework Equations

|\vec{v}|=\sqrt{0^2+0^2+20^2}=20
|\vec{B}|=\sqrt{1^2+1^2+0^2}=\sqrt{2}

The Attempt at a Solution

I have found the magnitude as 20\sqrt{2} which I'm pretty sure is right. My main problem is the new angle.
For the angle I have sin\theta=\frac{|\vec{v} \times \vec{B}}{|\vec{v}||\vec{B}|}=\frac{20\sqrt{2}}{20\sqrt{2}}
So the angle should be sin^{-1}(\frac{20\sqrt{2}}{20\sqrt{2}})
But the angle listed as the answer is 135 degrees.
Solving backwards, |\vec{v}||\vec{B}| would need to be 40 and I can't see where this comes from. Any input is greatly appreciated.

 

[Dick]

Homework Statement

Find the magnitude of \vec{v} \times \vec{B} and the direction of the new vector.
\vec{v}=\left \langle 0,0,20 \right \rangle
\vec{B}=\left \langle 1,1,0 \right \rangle

Homework Equations

|\vec{v}|=\sqrt{0^2+0^2+20^2}=20
|\vec{B}|=\sqrt{1^2+1^2+0^2}=\sqrt{2}

The Attempt at a Solution

I have found the magnitude as 20\sqrt{2} which I'm pretty sure is right. My main problem is the new angle.
For the angle I have sin\theta=\frac{|\vec{v} \times \vec{B}}{|\vec{v}||\vec{B}|}=\frac{20\sqrt{2}}{20\sqrt{2}}
So the angle should be sin^{-1}(\frac{20\sqrt{2}}{20\sqrt{2}})
But the angle listed as the answer is 135 degrees.
Solving backwards, |\vec{v}||\vec{B}| would need to be 40 and I can't see where this comes from. Any input is greatly appreciated.

Yes, the magnitude is 20\sqrt{2}. Though I'm not really sure how you got that. But what angle are you looking for? The angle you are computing is the angle between the two vectors which is 90 degrees. I think they are looking for the angle the new vector makes with respect to the x-axis. To answer I think you need to actually calculate the vector $\vec{v} \times \vec{B}$.

 

[HallsofIvy]

The direction of a vector in three dimensions cannot be given by a single angle.

Is "|u x v|= |u||v|sin(\theta)" and "u x v is perpendicular to both u and v by the right hand rule" the only way you have to find u x v? If u= ai+ bj+ ck and v= di+ ej+ fk then u x v= (bf- ce)i+ (ce- af)j+ (ae- bd)k is often simpler to use.