Why does cross product give a vector which is perpendicular to a plane I was wondering if anyone could give me the intuition behind the cross product of two vectors $\textbf{a}$ and $\textbf{b}$. Why does their cross product $\textbf{n} = \textbf{a} \times \textbf{b}$  give me a vector which is perpendicular to a plane?

I know I can just check this by using dot product but I'm not totally satisfied with "it just works" answer =)

Thank you for any help! =)
The determinant formula isn't so mysterious. Consider the cross product $\mathbf{v} = \langle a,b,c \rangle \times \langle d,e,f \rangle$ as the formal determinant

$$ \det \left(\begin{array}{ccc} \mathbf{i} &amp; \mathbf{j} &amp; \mathbf{k}\\ a &amp; b &amp; c \\ d &amp; e &amp; f \end{array} \right) $$

where $\mathbf{i}, \mathbf{j}, \mathbf{k}$ are the standard basis vectors. If instead one considers $\mathbf{i}, \mathbf{j}, \mathbf{k}$ as indeterminates and substitutes $x, y, z$ for them, this determinant computes the dot product $\mathbf{v} \cdot \langle x, y, z \rangle$. But letting $\langle x, y, z \rangle$ be $\langle a, b, c \rangle$ or $\langle d, e, f \rangle$ gives a zero determinant, so $\mathbf{v}$ is perpendicular to the latter two vectors, hence to the plane they span, as Omnomnomnom says.
I would start by something simpler.
What is i x j ?
answer: k
but why?  Well, it's by definition.  The cross product of a unit vector in the x-direction (i) and a unit vector in the y-direction (j) is a perpendicular vector in the z-direction  (k).
Given the above, one can easily see that:
2i x j = 2k
In 2 dimensions, if you want to multiply a x b

Simply breakdown vectors a and b into their i and j components.
Multiply the i from the first vector (a) with the j in the second vector (b).
Multiply the j from the first vector (a) with the i in the second vector (b).
Add them together, remembering that j x i is NEGATIVE k.

so step 2 and 3 work opposite of each other - if you will.
That's it.  Now think about this in 3 dimensions remembering that:
j x k = i
k x i = j
and that the reverse cross product gives a negative sign.