Why is the cross product perpendicular?

[JizzaDaMan]

Why is the cross product of two vectors perpendicular to the plane the two vectors lie on?

I am aware that you can prove this by showing that:

(\vec{a}\times\vec{b})\cdot\vec{a} = (\vec{a}\times\vec{b})\cdot\vec{b} = 0

Surely it was not defined as this and worked backwards though. I see little advantage in making this definition, and simply guessing it seems a bit random, so what brings it about?

 

[micromass]

What is your definition of the cross product?

 

[Ferramentarius]

By the matrix definition of the cross product we have
\vec{a}\times \vec{b} \cdot \vec{c} <br /> = \begin{vmatrix} \vec{i} &amp; \vec{j} &amp; \vec{k} \\ a_i &amp; a_j &amp; a_k \\ b_i &amp; b_j &amp; b_k \end{vmatrix} \cdot \vec{c}<br /> = (\vec{i} \begin{vmatrix} a_j &amp; a_k \\ b_j &amp; b_k \end{vmatrix} -\vec{j} \begin{vmatrix} a_i &amp; a_k \\ b_i &amp; b_k \end{vmatrix} + \vec{k} \begin{vmatrix} a_i &amp; a_j \\ b_i &amp; b_j \end{vmatrix} ) \cdot \vec{c} \\ <br /> = (c_i \begin{vmatrix} a_j &amp; a_k \\ b_j &amp; b_k \end{vmatrix} -c_j \begin{vmatrix} a_i &amp; a_k \\ b_i &amp; b_k \end{vmatrix} + c_k \begin{vmatrix} a_i &amp; a_j \\ b_i &amp; b_j \end{vmatrix} )<br /> = \begin{vmatrix} c_i &amp; c_j &amp; c_k \\ a_i &amp; a_j &amp; a_k \\ b_i &amp; b_j &amp; b_k \end{vmatrix}.

When \vec{c} = \vec{a} or \vec{c} = \vec{b} the determinant has two equal rows and becomes zero. This means the dot product is zero and the vectors are perpendicular.

 

[lurflurf]

The cross product is the (up to multiplication by a constant) only product possible that takes two vectors to a third. It is also extremely useful to produce a vector perpendicular to two given vectors. All the time you have two vectors and need one perpendicular to them. Bam! Cross product done.