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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.

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  • $\begingroup$ What intuition do you have for the cross product of two non-parallel vectors? $\endgroup$ May 16, 2017 at 18:29
  • $\begingroup$ Please do not delete questions which for which people have taken the time to provid answers unless there is a serious reason. $\endgroup$ May 16, 2017 at 19:24

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

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

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  • $\begingroup$ This is such a good answer. Way better than the accepted one! $\endgroup$
    – user3180
    Jan 18, 2021 at 11:57
  • $\begingroup$ Can you explain why $\| b \|$ disappears? $\endgroup$
    – user3180
    Jan 18, 2021 at 11:58
  • $\begingroup$ Nevermind - I should have downvoted this - the formula is a TYPO! $\endgroup$
    – user3180
    Jan 18, 2021 at 12:27
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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 $$\mathbf u\cdot \mathbf v = \|\mathbf v\|\|\operatorname{proj}_{\mathbf v}(\mathbf u)\|\mathbf n$$

Answer

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 $0$.

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  • $\begingroup$ I can go more in depth if needed. Just let me know. $\endgroup$
    – user137731
    May 18, 2017 at 1:09
  • $\begingroup$ That is perfect, thanks $\endgroup$
    – user296950
    May 18, 2017 at 6:53
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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 $2$

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The magnitude of the cross product is the same as the magnitude of one of them, multiplied by the component of one vector that is perpendicular to the other. If the vectors are parallel, no component is perpendicular to the other vector. Hence, the cross product is 0 although you can still find a perpendicular vector to both of these. You can see this for yourself by drawing 2 vectors 'a' and 'b', with an acute angle 'x' between the 2 vectors. To find the magnitude of the component of B that is perpendicular to A, it is Bsinx. And for the component of A that is perpendicular to B, it is also Bsinx.

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  • $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Mar 24, 2022 at 14:02

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