Vectors having the the same magnitude?

Question

Two vectors A and B each have magnatudes of 6.0 m and the angle between their directions is 60 degrees. Find AxB.
I know this is probably easy but I am still not sure.

Answer 1

This is fairly simple, construct a vector triangle and realise that the internal angle is 120degrees.

Then use the cosine rule to find the magnitude of the resultant vector:

a^2 = b^2 + c^2 - 2bc Cos(A)

A = 120
b= c =6

so, a^2 = 36+36 - 2(36)Cos(120)

a^2 = 108
a = 10.3923048...
a = 10.4 m (3 significant figures)

Since the triangle is essentially isosceles, the angle of the resultant vector is 30degrees from the first. (i.e. ((180 - 120)/ 2) )1

Answer 2

You are describing the cross product. The magnitude of the cross product is given by:

|| A X B || = || A || || B || sinθ
where θ is the angle between the vectors

In particular:

|| A X B || = || 6 || || 6 || sin 60° = 6*6*(√3/2) = 18√3

This is the magnitude of the cross product. The cross product itself is a vector of magnitude 18√3 that is perpendicular to the plane containing both of the given vectors.

Answer 3

If I remember right, AxB

magA*maB*sin theta

mag = magnitude of

theta = angle between 2 vectors.

Resultant vector will be normal to the plane of the 2 crossed vectors