# What Is The Magnitude Of The Cross Product Of Two Parallel Vectors? ?

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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\$\$ |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

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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\$.

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.

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\$\$