Vector A is 10 unit longs and points at 25 degree above the positive x axis. Another vector B is 8 units long and points along yhe positive x axis. Determine the magnitude of A X B.
If I'm reading this correctly, the question wants you to find the magnitude of vector A and then multiply it to magnitude of vector B. If that's the case then…
Vector A has both an x and y component:
Fx=10cos25=9.06
Fy=10sin25=4.266
Magnitude of A = sqrt(9.06^2+4.266^2)^(1/2) = 3.16
Vector B only has x components:
Fx=8
Magnitude of B = 8
3.16*8 = 25.28
Sorry, I made an error…
Vector A has both an x and y component:
Fx=10cos25=9.06
Fy=10sin25=4.266
Magnitude of A = sqrt(9.06^2+4.266^2) = 10.0141
Vector B only has x components:
Fx=8
Magnitude of B = 8
10.0141*8 = 80.112
To determine the magnitude of the cross product (A x B) of two vectors A and B, you can use the following formula:
|A x B| = |A| * |B| * sin(theta)
Where |A| represents the magnitude (length) of vector A, |B| represents the magnitude of vector B, and theta represents the angle between the two vectors.
In this case, we are given the length of vector A as 10 units. Additionally, vector B is given to be pointing along the positive x-axis, so the angle between vectors A and B is 90 degrees (since the positive x-axis is perpendicular to any vector pointing along it).
Therefore, the formula becomes:
|A x B| = 10 * 8 * sin(90)
Since sin(90) is equal to 1, the formula simplifies to:
|A x B| = 10 * 8 * 1
Evaluating this expression, we find:
|A x B| = 10 * 8 = 80
So, the magnitude of vector A x B is 80 units.