Question

Displacement 1 is in the yz plane 64.9o from the positive direction of the y axis, has a positive z component, and has a magnitude of 2.06 m. Displacement 2 is in the xz plane 30.7o from the positive direction of the x axis, has a positive z component, and has magnitude 0.870 m. What are (a) 12, (b) the x component of 1 x 2, (c) the y component of 1 x 2, (d) the z component of 1 x 2, and (e) the angle between 1 and 2?
Give your answers for (a) through (d) in standard SI units.

Answer 1

Okay, first break this down into components, I don't feel like crunching numbers... Also, I'm gonna call displacement 1 vector A and displacement 2 vector B.

\[A = 0x + |A|\cos(\theta_A)y+|A|\sin(\theta_A)z\]
\[B = |B|\cos(\theta_B)x + 0y + |B|\sin(\theta_B)z\]

The x, y, and z, after the stuff means in that direction.

1. Assuming this means the dot product. Just do
\[A*B = 0*|B|\cos(\theta_B) + |A|\cos(\theta_A)*0+|A|\sin(\theta_A)*|B|\sin(\theta_B)

2. For the cross product do
\[|A|*|B|\sin(\phi)\]

where phi is the angle between them. Then you'll have to do the right hand rule to figure out the direction and use trig to find the x component. It's a pain.

The rest of the questions follow the same pattern. It's mostly just alot of trig after that.