Why does cross product give a vector which is perpendicular to a plane

Question

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! =)

Answer 1

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

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

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.

Answer 2

See what happens when you try to take $(a\times b)\cdot a$ or $(a\times b)\cdot b$ (you should get $0$). If a vector is perpendicular to a basis of a plane, then it is perpendicular to that entire plane. So, the cross product of two (linearly independent) vectors, since it is orthogonal to each, is orthogonal to the plane which they span.

Also, while you're trying to develop an intuition for cross products, I highly recommend this video

https://www.khanacademy.org/math/linear-algebra/vectors_and_spaces/dot_cross_products/v/proof--relationship-between-cross-product-and-sin-of-angle

and, while we're there, might be worth knowing how the angle formula for dot products comes from the law of cosines

https://www.khanacademy.org/math/linear-algebra/vectors_and_spaces/dot_cross_products/v/defining-the-angle-between-vectors

Answer 3

My answer is that this is the case because the cross product is defined to have that property.

Answer 4

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

1. Simply breakdown vectors a and b into their i and j components.
2. Multiply the i from the first vector (a) with the j in the second vector (b).
3. Multiply the j from the first vector (a) with the i in the second vector (b).
4. 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.