B Confusion about the angle between two vectors in a cross product

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The magnitude of the cross product of two vectors A and B is given by |A×B| = |A||B|sinθ, where θ is the angle between the vectors defined within the range 0≤θ≤π to ensure a positive magnitude. When considering vectors A = 1i and B = -j, the angle can be measured as -π/2 clockwise or π/2 anti-clockwise, leading to confusion about which angle to use. The formula for the magnitude does not account for direction, focusing solely on the absolute value of sinθ, which is why θ should be treated as a non-negative angle. Thus, the angle used in the cross product formula is always taken as the positive equivalent within the defined range, ignoring the sign of the angle.
tbn032
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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.
 
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tbn032 said:
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.
We have ##\vec A \times \vec B = |A||B|\sin \theta##, hence ##|\vec A \times \vec B| = |A||B||\sin \theta|##. Or, when dealing with the modulus you could change your definition of ##\theta## to lie in the range ##[0, \pi)##.
 
The formula shown here (https://en.wikipedia.org/wiki/Cross_product, in the Properties section) gives the magnitude of the cross product as $$||\vec a \times \vec b|| = ||\vec a|| ||\vec b|| |\sin(\theta)|$$

BTW, in what you wrote there are some extra symbols whose purpose I don't understand.

But if A⃗=1i^ and B⃗ =-j^
What is the meaning of the square after A?
What is the meaning of the caret after j?
 
Mark44 said:
BTW, in what you wrote there are some extra symbols whose purpose I don't understand.What is the meaning of the square after A?
What is the meaning of the caret after j?
I just meant i cap and j cap (unit vectors along x-axis and y-axis, respectively)
 
PeroK said:
Or, when dealing with the modulus you could change your definition of ##\theta## to lie in the range ##[0, \pi)##.
I think the theta will lie in the range ##[0, \pi]## instead of ##[0, \pi)##.
 
tbn032 said:
I just meant i cap and j cap (unit vectors along x-axis and y-axis, respectively)
Please have a look at the "LaTeX Guide" link below the Edit window. That will help you to post your math equations in a much more readable form. Thank you.

1662143597191.png
 
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Note that the cross-product is often associated with the oriented parallelogram formed by its factors. (It might help to think "bi-vector. https://en.wikipedia.org/wiki/Bivector )
The magnitude of the cross-product is equal to the area of that parallelogram.
With the tails of the vectors together, the interior angle is a signed-angle whose magnitude is not larger than ##\pi##.
 
robphy said:
The magnitude of the cross-product is equal to the area of that parallelogram.
With the tails of the vectors together, the interior angle is a signed-angle whose magnitude is not larger than ##\pi##.
In many of the definition of cross product, I have seen the ##\theta##(angle between the two vector) is in the range of 0≤##\theta##≤π.

The angle between the two vector can be measured in two ways, clockwise and anti-clockwise. The clockwise wise measurement is generally taken to be negative and the anti-clockwise wise measurement is generally taken to be positive.

How can the ##\theta## always lie in the 0≤##\theta##≤π.for example, take two vectors ##\vec A## =1##\hat i## and ##\vec B## =-1##\hat j##.the angle between these vectors could be measured -π/2 and π/2 if we measure it clockwise and anticlockwise respectively. How can the -π/2 incorporated in the range 0≤##\theta##≤π.

is it the case that we ignore clockwise measurement when measuring the angle between vector or is it the case that the ##\theta## which is present in the formula |A⃗×B⃗|=|A⃗||B⃗|sin##\theta## is just the magnitude of the angle present between the two vector and direction(clockwise or anti-clockwise) is not considered(##\theta##=|angle between the vectors|)
 
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