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That is, how to prove the following identity: $$a \times (b+c) = a \times b + a \times c$$ where the $\times$ represents cross product of two vectors in 3-dimensional Euclidean space.

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4 Answers 4

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To see it geometrically:

Recall that $\lVert x\times y\rVert$ represents the area of the parallelogram with sides $x,y$. If you then "glue" together the parallelograms with sides $a,b$ and $a,c$ along the side $a$, you get a hexagon (not regular) with the same area of the parallelogram with sides $a,b+c$.

Sum of parallelograms.


Edit: As another user correctly noted, the above proof works only when $a$ lies in the plane spanned by $b$ and $c$. If not, then we can reduce to this case by considering the triangle $T$ with basis $b+c$ and sides $b,c$, as suggested by this image1:

sum of bivectors

Indeed, without loss of generality we may assume that $a$ is orthogonal to $b$ and $c$. Then the angle and proportion between $a \times b$ and $a \times c$ are the same as those between $b$ and $c$, because in this case the cross product with $a$ is equivalent to the composition of a rotation by $90$ degrees and a dilation by $\lVert a \rVert$. Therefore if $P$ is the prism with base $T$ and height $a$, then the area of the projection of the face $[a,b]$ on the face $[a,b+c]$ is the same as the length of the cross product between $a$ and the projection of $b$ on $b+c$, and similarly for the face $[a,c]$.

1. Courtesy of WikiMedia.

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  • $\begingroup$ I added a descriptive image. $\endgroup$
    – A.P.
    Apr 15, 2013 at 17:12
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    $\begingroup$ Could whoever downvoted this explain why they did so, allowing me to improve my answer? $\endgroup$
    – A.P.
    Apr 5, 2015 at 12:26
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    $\begingroup$ Your answer assumes that $a$, $b$ and $c$ all lie in the same plane. When they don't, which they don't in general, it breaks down. $\endgroup$ Apr 7, 2015 at 12:51
  • $\begingroup$ ($a \times b$ and $a \times c$ are no longer parallel, so the magnitude of their sum is less than the sum of their magnitudes; the picture at the bottom fails because it's now depicting a 3-d solid rather than a planar shape.) $\endgroup$ Apr 7, 2015 at 12:53
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    $\begingroup$ @user73985 thanks for catching the omission. I amended my answer and it should be fine now. $\endgroup$
    – A.P.
    Apr 10, 2015 at 19:28
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Just to enrich the post for future readers, I would like to add another derivation that I found on the internet. This proof uses the distributivity of the dot product (which is easier to prove), and the property that the circular commutation of vectors doesn't change the triple product of the vectors (which is quite obvious, since the triple product is just the volume of the parallelepiped formed by the vectors).

Let $d = a \times (b + c) - a \times b - a \times c$

so it is required to prove that $d = 0$:

$d^2$ = $d \cdot d$

$= d \cdot (a \times (b + c) - a \times b - a \times c)$

$= d \cdot (a \times (b + c)) - d \cdot (a \times b) - d \cdot (a \times c)$

$= (d \times a) \cdot (b + c) - (d \times a) \cdot b - (d \times a) \cdot c$

$= (d \times a) \cdot (b + c) - (d \times a) \cdot (b + c)$

$= 0$

Therefore $d = 0$, so $a \times (b + c) = a \times b + a \times c$.

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    $\begingroup$ I think you need to add a bit more work for this proof to be convincing. Namely, prove that (i) the triple product is the volume of the parallelepiped; and (ii) the triple product is circularly commutative without relying on the distributivity of the cross product. Otherwise one wonders if these latter two facts somehow depend on the fact that the cross product is distributive. (I know they don't, but there is this gap in your answer.) $\endgroup$
    – user46234
    Apr 22, 2016 at 5:15
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Let $a=(a_1,a_2,a_3),~~b=(b_1,b_2,b_3),~~c=(c_1,c_2,c_3)\in\mathbb R^3$ so $$a\times(b+c)=\begin{vmatrix} i\;\;\;j\;\;\;k \\ a_1\;\;\;a_2\;\;\;a_3 \\ b_1+c_1\;\;\;b_2+c_2\;\;\;b_3+c_3 \end{vmatrix}=i\left(a_2b_3+a_2c_3-a_3b_2-a_3c_2\right)-j(...)+k(...)$$ Now try to rearrange the above terms to find the result. See that in the first term we have $i\left(a_2b_3+a_2c_3-a_3b_2-a_3c_2\right)=i(a_2b_3-a_3b_2)+i(a_2c_3-a_3c_2)$.

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    $\begingroup$ I think, in this proof, the conclusion is implicitly assumed. $\endgroup$
    – Berci
    Apr 15, 2013 at 9:25
  • $\begingroup$ @Berci: As you commented before, I don't know what the OP know and how he/she wanted to do the problem. $\endgroup$
    – Mikasa
    Apr 15, 2013 at 9:27
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    $\begingroup$ Doesn't this expression (as a determinant) for the cross product of two vectors itself require distributivity of the cross product? $\endgroup$ Apr 22, 2018 at 10:47
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Found a geometric proof in https://en.wikiversity.org/wiki/Cross_product. Refer to section: Equivalence of the two Definitions

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    $\begingroup$ The question has already an accepted answer. I'm not sure your answer is highly useful, in my opinion. $\endgroup$
    – Watson
    Feb 4, 2016 at 13:29
  • $\begingroup$ This. The proof is straightforward if you project a and b onto the plane P orthogonal to c. $\endgroup$ Oct 24, 2019 at 19:17
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    $\begingroup$ This should be the accepted answer. $\endgroup$ Oct 24, 2019 at 19:37

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