Evaluate the cross product A⃗ ×B⃗ 𝐴→×𝐵→ of the two vectors above. They are separated by an angle of 150.° and have magnitudes given by: Vector A: 9.00 Vector B: 6.00
Question
Evaluate the cross product A⃗ ×B⃗ 𝐴→×𝐵→ of the two vectors above. They are separated by an angle of 150.° and have magnitudes given by: Vector A: 9.00 Vector B: 6.00
Solution
The cross product of two vectors A and B is given by the formula:
A x B = |A| * |B| * sin(θ) * n
where:
- |A| and |B| are the magnitudes of vectors A and B respectively,
- θ is the angle between A and B,
- n is the unit vector perpendicular to the plane containing A and B.
Given:
- |A| = 9.00
- |B| = 6.00
- θ = 150°
We can substitute these values into the formula:
A x B = 9.00 * 6.00 * sin(150°) * n
To find the sin(150°), you can use a calculator or a trigonometric table. The sin(150°) is 0.5.
So,
A x B = 9.00 * 6.00 * 0.5 * n A x B = 27.00 * n
Therefore, the cross product of vectors A and B is 27.00n. The direction of this vector (given by n) is perpendicular to the plane containing A and B. The exact direction (whether it points 'upwards' or 'downwards') depends on the 'right-hand rule'.
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