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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.

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    $\begingroup$ just take $\theta = \frac{π}{2}$. draw a diagram of the two vectors and you'll understand why. $\endgroup$
    – user857163
    Commented Sep 1, 2022 at 22:33
  • $\begingroup$ 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. $\endgroup$
    – hsdfasd
    Commented Sep 2, 2022 at 13:57

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$\theta$ can be measured in either a clockwise or counterclockwise direction, and you want to take the positive value of the smaller of the two angles. Your $\theta=-\frac{\pi}{2}$ should be plugged into the formula as $\frac{\pi}{2}$. The difference will manifest itself in the direction of the normal vector, $\vec{n}$, which is the resultant of the cross product. The rule of thumb is: "counterclockwise up, clockwise down."

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