How do I do the cross product of the following (<-2sin(t),2cos(t),2>/√8 cross(<-2cos(t),-2sin(t),0>/√4)?

3 answers

the same as any other two vectors. As a determinant. Just to make things easier to read, I'll save the √8 and √4 factors till the end.
|i j k|
| -2sint 2cost 2|
|-2cost -2sint 0|
= <4sint,-4cost,8sint cost>
or, 4<sint, -cost, sin2t>
so, since 4/(√8√4) = 1/√2,
1/√2 <sint, -cost, sin2t>
I really need those steps.Thank you
In that case, review your Algebra II, where you learned to evaluate 3x3 determinants.

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