- #1

togame

- 18

- 0

## Homework Statement

Find the magnitude of [itex]\vec{v} \times \vec{B}[/itex] and the direction of the new vector.

[itex]\vec{v}=\left \langle 0,0,20 \right \rangle[/itex]

[itex]\vec{B}=\left \langle 1,1,0 \right \rangle[/itex]

## Homework Equations

[itex]|\vec{v}|=\sqrt{0^2+0^2+20^2}=20[/itex]

[itex]|\vec{B}|=\sqrt{1^2+1^2+0^2}=\sqrt{2}[/itex]

## The Attempt at a Solution

I have found the magnitude as [itex]20\sqrt{2}[/itex] which I'm pretty sure is right. My main problem is the new angle.

For the angle I have [itex]sin\theta=\frac{|\vec{v} \times \vec{B}}{|\vec{v}||\vec{B}|}=\frac{20\sqrt{2}}{20\sqrt{2}}[/itex]

So the angle should be [itex]sin^{-1}(\frac{20\sqrt{2}}{20\sqrt{2}})[/itex]

But the angle listed as the answer is 135 degrees.

Solving backwards, [itex]|\vec{v}||\vec{B}|[/itex] would need to be 40 and I can't see where this comes from. Any input is greatly appreciated.