- #1

- 34

- 7

The magnitude of cross product is defined of vector A⃗ and B⃗ as |A⃗×B⃗|=|A⃗||B⃗|sinθ where θ is defined as the angle between the two vector and 0≤θ≤π.the domain of θ is defined 0≤θ≤π so that the value of sinθ remains positive and thus the value of the magnitude |A⃗||B⃗|sinθ also remain positive (magnitude cannot be negative).

But if A⃗=1i^ and B⃗ =-j^ then the angle between these vector would be -π/2 but the magnitude of |A⃗×B⃗|=|A⃗||B⃗|sinθ where the domain of θ is defined 0≤θ≤π.

how can this angle(-π/2) be incorporated in the formula so that the magnitude of cross product of these vectors could be found. In general, how can the θ whose value is π≤θ≤2π incorporated in the formula so that the magnitude of the cross product of the vectors can be calculated.

The angle between vector A⃗ and B⃗ can be θ=π/2 if we measure anti-clockwise (From vector B⃗ to A⃗ and θ=-π/2 if we measure the angle clockwise(from vector A⃗ to B⃗).how can I say which angle to pick.

But if A⃗=1i^ and B⃗ =-j^ then the angle between these vector would be -π/2 but the magnitude of |A⃗×B⃗|=|A⃗||B⃗|sinθ where the domain of θ is defined 0≤θ≤π.

how can this angle(-π/2) be incorporated in the formula so that the magnitude of cross product of these vectors could be found. In general, how can the θ whose value is π≤θ≤2π incorporated in the formula so that the magnitude of the cross product of the vectors can be calculated.

The angle between vector A⃗ and B⃗ can be θ=π/2 if we measure anti-clockwise (From vector B⃗ to A⃗ and θ=-π/2 if we measure the angle clockwise(from vector A⃗ to B⃗).how can I say which angle to pick.

Last edited by a moderator: