Everyone of us know about the vector cross product. But I wonder, how the formula of $AB\sin\theta$ has been derived? Can anyone help?

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    $\begingroup$ The given question explains well the dot product, but not the cross product $\endgroup$
    – Aneek
    Commented Aug 16, 2015 at 11:05
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    $\begingroup$ The question is duplicate. $\endgroup$ Commented Aug 16, 2015 at 11:07
  • $\begingroup$ I have not intended to duplicate it, I just want to know the answer to it $\endgroup$
    – Aneek
    Commented Aug 16, 2015 at 11:08
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    $\begingroup$ @Aneek; Hi, do you know what duplicate is?? You have every right to know the answer but we can't simply answer the same question again & again but fortunately you've got a pretty good answer from @Alfred Centuarui. My suggestion will be that first google your query; many of your queries were trivial i.e. they could be easily found by googling. If problem persists, then come here but at first check if it had been asked before or some related question has been asked before. If they answer your query, then you ought not ask that because then it would be marked as duplicate. Hope this helps. $\endgroup$
    – user36790
    Commented Aug 16, 2015 at 14:31
  • $\begingroup$ BTW, I'm removing the downvote since you are now here & I'm seeing you still need some more time to acclimatize here:) $\endgroup$
    – user36790
    Commented Aug 16, 2015 at 14:33

1 Answer 1


One can define the (magnitude) of the cross product this way or better

$$\mathbf A \times \mathbf B = AB\sin\theta\; \mathbf n $$

where $\mathbf n$ is the (right hand rule) vector normal to the plane containing $\mathbf A$ and $\mathbf B$,

Another approach is to start by specifying the cross product on the Cartesian basis vectors:

$$\vec e_x \times \vec e_y = \vec e_z = -(\vec e_y \times \vec e_x)$$

$$\vec e_y \times \vec e_z = \vec e_x = -(\vec e_z \times \vec e_y)$$

$$\vec e_z \times \vec e_x = \vec e_y = -(\vec e_x \times \vec e_z)$$

Or, more succinctly:

$$\vec e_i \times \vec e_j = \epsilon_{ijk}\vec e_k$$

Now, we can always orient the coordinate system such that the vectors $\mathbf A$ and $\mathbf B$ are in the $xy$ plane.


$$\mathbf A \times \mathbf B = (A_x \vec e_x + A_y \vec e_y) \times (B_x \vec e_x + B_y \vec e_y) = (A_xB_y - A_yB_x)\vec e_z$$

The final term in parenthesis can be written as:

$$(A\cos\theta_a\;B\sin\theta_b - A\sin\theta_a\;B\cos\theta_b) = AB(\cos\theta_a\sin\theta_b - \sin\theta_a\cos\theta_b)$$

$$= AB\sin(\theta_b - \theta_a) = AB \sin \theta$$

where $\theta = \theta_b - \theta_a$ is the angle counter-clockwise (right-hand rule) from $\mathbf A$ to $\mathbf B$ .

Since the lengths $A$ and $B$ of, and the angle $\theta$ between, the vectors are coordinate independent quantities, this is a coordinate independent result:

$$||\mathbf A \times \mathbf B|| = AB\sin\theta $$


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