## Why is the cross product perpendicular?

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

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## Why is the cross product perpendicular?

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.
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