What is the intuition behind the cross product of two parallel vectors being zero?

I understand why the cross product, c, of two parallel vectors, a & b, is zero both from the definition that ||c|| = ||a|| ||b|| sin(x) (and sin(0) = 0), and from the component wise calculation of the cross product.

However, I don't see the intuition behind why. Surely if a and b are parallel it is still possible to find a third vector, c, that is orthogonal to these two? What am I missing?

Although this post is very useful for explaining the logic behind cross products What is the logic/rationale behind the vector cross product?, it didn't contain enough relevant material as to the intuition behind why the cross product of two parallel vectors is equal to zero.

• What intuition do you have for the cross product of two non-parallel vectors? – Mariano Suárez-Álvarez May 16 '17 at 18:29
• Please do not delete questions which for which people have taken the time to provid answers unless there is a serious reason. – Mariano Suárez-Álvarez May 16 '17 at 19:24

Build-Up

One intuitive way to define/ characterize the dot and cross products is in terms of the projection and rejection operations.

The dot product can be thought of as a way to measure the length of the projection of a vector $\mathbf u$ onto a vector $\mathbf v$. Specifically $$\mathbf u\cdot \mathbf v = \pm\|\mathbf v\|\|\operatorname{proj}_{\mathbf v}(\mathbf u)\|$$ where the sign is determined by whether the angle between the vectors is acute or obtuse. And of course, when the angle between the vectors is right, the projection $\operatorname{proj}_{\mathbf v}(\mathbf u)$ is just the zero vector $\mathbf 0$, and hence $\mathbf u\cdot \mathbf v = 0$.

The cross product, analogously, can be thought of as a way to measure the length of the rejection of a vector $\mathbf u$ from a vector $\mathbf v$. However, if you also want information on which direction the rejection points in (which is analogously given by the $\pm$ in the dot product), then it turns out that there's no way to do this consistently in a three dimensional space using a scalar. But we can describe the direction using a vector quantity -- though to make it a bit more symmetric (or technically anti-symmetric) in the arguments of the cross product, there are only two choices that we could make for the direction (they correspond to the usual right-hand rule and an alternate way of defining the cross product via a left-hand rule).

I won't go through the full process of motivating and constructing the cross product here as it's not necessarily a short argument and besides it's probably something that you'll learn better by working it out on your own. But here's the way we define/ characterize the cross product in $\Bbb R^3$ $$\mathbf u\times \mathbf v = \|\mathbf v\|\|\operatorname{rej}_{\mathbf v}(\mathbf u)\|\mathbf n$$ where $\mathbf n$ is the unique unit vector in our (three dimensional) space which is orthogonal to both $\mathbf u$ and $\mathbf v$ and where $(\mathbf u,\mathbf v, \mathbf n)$ forms a right-handed sequence -- i.e. the unit vector we obtain from the right-hand rule.

To further expand on the relationship between the definitions of the dot and cross products, notice that in the vector space $\Bbb R$, the only unit "vectors" are $1$ and $-1$. So if we defined $\mathbf n$ to be $1$ when the angle between $\mathbf u$ and $\mathbf v$ is acute and $-1$ otherwise, then we could write $$\mathbf u\cdot \mathbf v = \|\mathbf v\|\|\operatorname{proj}_{\mathbf v}(\mathbf u)\|\mathbf n$$

So the answer to your question is that the cross product of two parallel vectors is $\mathbf 0$ because the rejection of a vector from a parallel vector is $\mathbf 0$ and hence has length $0$.

• I can go more in depth if needed. Just let me know. – user137731 May 18 '17 at 1:09
• That is perfect, thanks – user296950 May 18 '17 at 6:53

$$\|\vec{a} \times \vec{b}\| = \|\vec{a}\| \sin \theta = \textbf{Area of Plane spanned by \vec{a}, \vec{b}}$$

Here $\theta$ is the angle between $\vec{a}, \vec{b}$ and so if $\vec{a}, \vec{b}$ are parallel, we have $\theta = 0 \Rightarrow \vec{a} \times \vec{b} = 0$

• This is such a good answer. Way better than the accepted one! – user3180 Jan 18 at 11:57
• Can you explain why $\| b \|$ disappears? – user3180 Jan 18 at 11:58
• Nevermind - I should have downvoted this - the formula is a TYPO! – user3180 Jan 18 at 12:27

You are presumably familiar with the fact the cross product is antisymmetric:

$$a \times b = - b \times a$$

By plugging in $b = a$, you can compute

$$a \times a = -a \times a$$

This is a basic intuition about antisymmetry — anytime you have something antisymmetric, something special happens with repeated arguments.

In the current setting, we can further compute

$$2(a \times a) = 0$$ $$a \times a = 0$$

thus the cross product is an alternating product. (that's what a product satisfying this last identity is called)

This is a basic intuition of multilinear algebra: "antisymmetric" and "same¹ thing for products. For completeness, let me show that you can derive antisymmetry from alternation:

$$(a+b) \times (a+b) = 0$$ $$(a \times a) + (a \times b) + (b \times a) + (b \times b) = 0$$ $$0 + (a \times b) + (b \times a) + 0 = 0$$ $$a \times b = - b \times a$$

1: Things are more complicated if you're in a more general setting where you can't divide by $2$