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$$\mathbf{a}=a_1\mathbf{i}+a_2\mathbf{j}+a_3\mathbf{k}$$ $$\mathbf{b}=b_1\mathbf{i}+b_2\mathbf{j}+b_3\mathbf{k}$$ $$\mathbf{c}=c_1\mathbf{i}+c_2\mathbf{j}+c_3\mathbf{k}$$

Use appropriate determinants to prove that $$(\mathbf{a}\times\mathbf{b})+(\mathbf{a}\times\mathbf{c})=\mathbf{a}\times(\mathbf{b}+\mathbf{c})$$

My answers for each side don't match up. Can someone please help?

MattAllegro
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Hasan
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  • possible duplicate: http://math.stackexchange.com/questions/362139/how-to-prove-the-distributive-property-of-cross-product – Emilio Novati Feb 20 '15 at 21:49

1 Answers1

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Using the determinants as usual, for the $\mathbf{i}$-component of $\mathbf{a}\times(\mathbf{b}+\mathbf{c})$ you have: $$ \mathbf{i}\left[a_2(b_3+c_3)-a_3(b_2+c_2) \right]= $$ $$ \mathbf{i}\left[a_2b_3+a_2c_3-a_3b_2-a_3c_2 \right]= $$ $$ \mathbf{i}\left[(a_2b_3-a_3b_2)+(a_2c_3-a_3c_2) \right]= $$ $$ \mathbf{i}\left[(a_2b_3-a_3b_2)\right]+\mathbf{i}\left[(a_2c_3-a_3c_2) \right] $$ that is the $\mathbf{i}$-component of $\mathbf{a}\times\mathbf{b}+\mathbf{a}\times\mathbf{c}$

And the same for other components.

Emilio Novati
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