Find u X v, where Q is the angle between u and v.
magnitude of u =4
magnitude of v=10
Q=2pi/3
The Q is the part that is throwing me off. I do not know where to start.
To find u X v, where Q is the angle between u and v, we can use the formula for the cross product in terms of magnitudes and the angle between them:
u X v = |u| * |v| * sin(Q) * n
where |u| is the magnitude of vector u, |v| is the magnitude of vector v, Q is the angle between u and v, and n is the unit vector perpendicular to the plane containing u and v.
Given that |u| = 4, |v| = 10, and Q = 2pi/3, we can substitute these values into the formula:
u X v = 4 * 10 * sin(2pi/3) * n
Now let's calculate sin(2pi/3). Recall that the sine function has a value of √3/2 at 2pi/3:
sin(2pi/3) = √3/2
Substituting this into the formula:
u X v = 4 * 10 * (√3/2) * n
Simplifying:
u X v = 20√3 * n
So the cross product of u and v is 20√3 times the unit vector n. Note that we do not have enough information to calculate the direction of the resulting vector, but we know it is perpendicular to the plane containing u and v.