Vector A and Vector B has magnitudes of 12 and 10, respectively. If the angle between these vectors is 250, what is the magnitude of the cross product of vectors A and B?
0
109
120
51
Answers
Explanation:
Let the two vectors be A and B
The magnitude of their cross product is given by |A||B|sin(theta) where theta is the angle between the two vectors.
Magnitude of their dot product is given by |A||B|cos(theta)
Therefore the sum of the squares of the magnitude of the cross product and the dot product must be equal to (|A||B|)^2
as [sin(theta)]^2+[cos(theta)]^2=1
So the magnitude of the dot product is
[(magnitude of A*magnitude of B)^2-(magnitude of cross product)^2]^0.5
or,
-[(magnitude of A*magnitude of B)^2-(magnitude of cross product)^2]^0.5
Answer:
Let the two vectors be A and B
The magnitude of their cross product is given by |A||B|sin(theta) where theta is the angle between the two vectors.
Magnitude of their dot product is given by |A||B|cos(theta)
Therefore the sum of the squares of the magnitude of the cross product and the dot product must be equal to (|A||B|)^2
as [sin(theta)]^2+[cos(theta)]^2=1
So the magnitude of the dot product is
[(magnitude of A*magnitude of B)^2-(magnitude of cross product)^2]^0.5
or,
-[(magnitude of A*magnitude of B)^2-(magnitude of cross product)^2]^0.5