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.

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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.

linear-algebra vectors intuition cross-product

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edited May 18 “17 at 1:11

user137731

asked May 16 “17 at 18:26

user296950user296950

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## 3 Answers 3

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$$ |vec{a} imes vec{b}| = |vec{a}| sin heta = extbf{Area of Plane spanned by $vec{a}, vec{b}$}$$

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

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answered May 16 “17 at 18:29

Faraad ArmwoodFaraad Armwood

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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 $rianowens.tvbf u$ onto a vector $rianowens.tvbf v$. Specifically $$rianowens.tvbf ucdot rianowens.tvbf v = pm|rianowens.tvbf v||operatorname{proj}_{rianowens.tvbf v}(rianowens.tvbf u)|$$ 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 $operatorname{proj}_{rianowens.tvbf v}(rianowens.tvbf u)$ is just the zero vector $rianowens.tvbf 0$, and hence $rianowens.tvbf ucdot rianowens.tvbf v = 0$.

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The cross product, analogously, can be thought of as a way to measure the length of the rejection of a vector $rianowens.tvbf u$ from a vector $rianowens.tvbf v$. 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 $Bbb R^3$ $$rianowens.tvbf u imes rianowens.tvbf v = |rianowens.tvbf v||operatorname{rej}_{rianowens.tvbf v}(rianowens.tvbf u)|rianowens.tvbf n$$ where $rianowens.tvbf n$ is the unique unit vector in our (three dimensional) space which is orthogonal to both $rianowens.tvbf u$ and $rianowens.tvbf v$ and where $(rianowens.tvbf u,rianowens.tvbf v, rianowens.tvbf n)$ forms a right-handed sequence — i.e. the unit vector we obtain from the right-hand rule.

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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 $rianowens.tvbf n$ to be $1$ when the angle between $rianowens.tvbf u$ and $rianowens.tvbf v$ is acute and $-1$ otherwise, then we could write $$rianowens.tvbf ucdot rianowens.tvbf v = |rianowens.tvbf v||operatorname{proj}_{rianowens.tvbf v}(rianowens.tvbf u)|rianowens.tvbf n$$

## Answer

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