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 a and b. Why does their cross product n=a×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! =)

Answer

The determinant formula isn’t so mysterious. Consider the cross product v=a,b,c×d,e,f as the formal determinant

det(ijkabcdef)

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.

Attribution
Source : Link , Question Author : jjepsuomi , Answer Author : Brendan Pawlowski

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