How do I do the cross product of the following (<-2sin(t),2cos(t),2>/√8 cross(<-2cos(t),-2sin(t),0>/√4)?
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.
google can provide you with discussions and videos, in much greater depth than we can provide here.
To calculate the cross product of two vectors, you can use the following formula:
(A × B) = (A_y * B_z - A_z * B_y)i + (A_z * B_x - A_x * B_z)j + (A_x * B_y - A_y * B_x)k
In this case, you have two vectors:
A = <-2sin(t), 2cos(t), 2> / √8
B = <-2cos(t), -2sin(t), 0> / √4
To calculate the cross product, you will perform the following steps:
Step 1: Assign values to A and B
A_x = -2sin(t) / √8
A_y = 2cos(t) / √8
A_z = 2 / √8
B_x = -2cos(t) / √4
B_y = -2sin(t) / √4
B_z = 0
Step 2: Calculate the cross product components
(A_y * B_z - A_z * B_y) = (2cos(t) / √8 * 0) - (2 / √8 * (-2sin(t) / √4))
(A_z * B_x - A_x * B_z) = (2 / √8 * (-2cos(t) / √4)) - (-2sin(t) / √8 * 0)
(A_x * B_y - A_y * B_x) = ((-2sin(t) / √8 * (-2sin(t) / √4)) - (2cos(t) / √8 * (-2cos(t) / √4)))
Step 3: Simplify the cross product components
(A_y * B_z - A_z * B_y) = -4sin(t) / 4
(A_z * B_x - A_x * B_z) = -4cos(t) / 8
(A_x * B_y - A_y * B_x) = 2sin^2(t) / 8 + 2cos^2(t) / 8
Step 4: Combine the cross products components
(A × B) = (-sin(t), -cos(t), sin^2(t) + cos^2(t)) / 2
So, the cross product of the given vectors (<-2sin(t), 2cos(t), 2>/√8) × (<-2cos(t), -2sin(t), 0>/√4) is (-sin(t), -cos(t), 1) / 2.