# Derivation of vector cross product [duplicate]

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?

• The given question explains well the dot product, but not the cross product – Aneek Aug 16 '15 at 11:05
• The question is duplicate. – Norbert Schuch Aug 16 '15 at 11:07
• I have not intended to duplicate it, I just want to know the answer to it – Aneek Aug 16 '15 at 11:08
• BTW, I'm removing the downvote since you are now here & I'm seeing you still need some more time to acclimatize here:) – user36790 Aug 16 '15 at 14:33

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.

Thus:

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

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:

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