Cross product of two sums of three vectors?

This seems like a simple question but I couldn't find anywhere to verify

Is it true that:

$$(\mathbf{a}+\mathbf{b}+\mathbf{c})\times (\mathbf{d}+\mathbf{e}+\mathbf{f}) = (\mathbf{a}\times \mathbf{d})+(\mathbf{a}\times\mathbf{e})+(\mathbf{a}\times\mathbf{f}) + (\mathbf{b}\times \mathbf{d})+(\mathbf{b}\times\mathbf{e})+(\mathbf{b}\times\mathbf{f}) + (\mathbf{c}\times \mathbf{d})+(\mathbf{c}\times\mathbf{e})+(\mathbf{c}\times\mathbf{f}) $$

Just wondering whether the following identity is true.

Solution 1:

This was already ansered in comments, so I'll just write it in answers from Colescu's comment. We know that the cross product is distributive, meaning $(a+b)\times c=a\times c + b \times c$ and $c×(a+b)=c×a+c×b$

We can do this a bunch of times from the left to arrive to the right side of the equation $(a+b+c)×(d+e+f)=(a×d)+(a×e)+(a×f)+(b×d)+(b×e)+(b×f)+(c×d)+(c×e)+(c×f)$

So we start with $(a+b+c)×(d+e+f)$, We'll keep using the distributive propiety until we arrive at the desired equation.

$$(a+b+c)×(d+e+f)$$ $$=(a+[b+c])×(d+e+f)$$ $$=(a×(d+e+f))+([b+c]×(d+e+f))$$ $$=(a×(d+e+f))+(b×(d+e+f))+(c×(d+e+f))$$ $$=(a×(d+[e+f]))+(b×(d+[e+f]))+(c×(d+[e+f]))$$ $$=(a×d+a×[e+f])+(b×d+b×[e+f])+(c×d+c×[e+f])$$ $$=(a×d+a×e+a×f)+(b×d+b×e+b×f)+(c×d+c×e+c×f)$$ $$=(a×d)+(a×e)+(a×f)+(b×d)+(b×e)+(b×f)+(c×d)+(c×e)+(c×f)$$