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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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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 $a,b+c$.

**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 image^{1}:

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.}

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 $i\left(a_2b_3+a_2c_3-a_3b_2-a_3c_2\right)=i(a_2b_3-a_3b_2)+i(a_2c_3-a_3c_2)$.

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 $a \times (b + c) = a \times b + a \times c$.

Found a geometric proof in https://en.wikiversity.org/wiki/Cross_product. Refer to section: Equivalence of the two Definitions

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