Respuesta :
Answer: Hi!
First, UxV = sin(a)*IUI*IVI
where a is the angle between U and V, in this case 45°.
First, the cross product of UxV points:
Here you can use the right hand method,
Put your hand flat, so the point of your fingers point in the same direction that the first vector, in this case U, so your fingers will point to the north.
Now roll your fingers in the direction of the second vector, so here you will roll your fingers in the northeast direction. Now you will see that your thumb is pointing down, so the cross product of UxV points down.
And the magnitude is 6*5*sin(45) = 21.213
The cross-product u×v points: Option D: Down, and its magnitude |u×v| evaluates to [tex]|u \times v| = 15\sqrt{2}[/tex]
How to find the cross product of two vectors?
Suppose that two vectors in consideration are u and v, then their cross product is evaluated as:
[tex]u \times v = |u|.|v|.sin(\theta)\hat{n}[/tex]
where [tex]\hat{n}[/tex] is the normal unit vector whose direction is decided by right hand thumb rule, and theta is the angle between u and v vector.
The two bars around a vector represents the magnitude of that vector.
Cross product returns the result as a vector itself.
For this case, we have:
- |u| = 6 points, its direction is in north
- |v| = 5 points, its direction is in northeast
Thus, as north and northeast have 45 degrees in between them, therefore, we get:
[tex]u \times v = 6\times 5 \times sin(45^\circ) \hat{n} = 15\sqrt{2} \: \hat{n}[/tex]
Directing index finger of right hand to north direction, and middle to northeast makes the thumb go down, therefore, the direction of normal vector (and therefore direction of the resultant cross product vector too) is downside of this whole north south east west plane.
The magnitude of cross product is [tex]|u \times v| = 15\sqrt{2}[/tex]
Thus, the cross-product u×v points: Option D: Down, and its magnitude |u×v| evaluates to [tex]|u \times v| = 15\sqrt{2}[/tex]
Learn more about cross product here:
https://brainly.com/question/1968115
