- #1
odie5533
- 58
- 0
Homework Statement
Given two vectors [tex]\vec{A}=(4.25m/s)\hat{_i}+(5.00m/s)\hat{_j}-(1.25m/s)\hat{_k}[/tex] and [tex]\vec{B}=-(3.75m/s)\hat{_j} + (0.75m/s)\hat{_k}[/tex], a) find the magnitude of each vector; b) write an expression for [tex]\vec{A} - \vec{B}[/tex]. (c) Find the magnitude and direction of [tex]\vec{A} \times \vec{B}[/tex].
The Attempt at a Solution
a)[tex]|\vec{A}| = \sqrt{44.625m/s}[/tex]
[tex]|\vec{B}| = \sqrt{14.625m/s}[/tex]
b)[tex]\vec{A} - \vec{B} = 4.25m/s\hat{_i} + 8.75m/s\hat{_j} - 2.00m/s\vec{_k}[/tex]
c)[tex]\vec{A} \times \vec{B} = (4.25m/s\hat{_i} + 5.00m/s\hat{_j} - 1.25m/s\hat{_k}) \times (-3.75m/s\hat{_j} + 0.75m/s\hat{_k})[/tex]
[tex]\vec{C} = \vec{A} \times \vec{B}[/tex]
[tex]\vec{C}_x = (5.00m/s \times 0.75m/s) - (-1.25m/s \times -3.75m/s) = -0.9375m/s[/tex]
[tex]\vec{C}_y = (-1.25m/s \times 0) - (4.25m/s \times 0.75m/s) = -3.1875m/s[/tex]
[tex]\vec{C}_z = (4.25m/s \times -3.75m/s) - (5.00m/s \times 0) = -15.9m/s[/tex]
[tex]\vec{C} = -0.9375m/s\hat{_i} - 3.1875m/s\hat{_j} - 15.9m/s\hat{_k}[/tex]
[tex]|\vec{A} \times \vec{B}| = |\vec{C}| = \sqrt{(-0.9375m/s)^2 + (-3.1875m/s)^2 + (-15.9m/s)^2} = 16.3m/s[/tex]
Direction of [tex]\vec{A} \times \vec{B}[/tex] = ?
I'm not sure how to define the direction of a 3D vector is what it really comes down to.