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If $\overrightarrow{a} = \langle a_1, a_2, a_3 \rangle$ and $\overrightarrow{b} = \langle b_1, b_2, b_3 \rangle$, then the cross product of $\overrightarrow{a}$ and $\overrightarrow{b}$ is the vector

$$\overrightarrow{a} \times \overrightarrow{b} =(a_2 b_3 - a_3 b_2)\overrightarrow{i} + (a_3 b_1 - a_1 b_3)\overrightarrow{j} + (a_1 b_2 - a_2 b_1)\overrightarrow{k}$$

Using the above definition show algebraically that

$$|\overrightarrow{a} \times \overrightarrow{b}|^2=|\overrightarrow{a}|^2|\overrightarrow{b}|^2-(\overrightarrow{a}\cdot\overrightarrow{b})^2$$

I'm not sure if I'm on the right track with this problem, but I've started with the given expression

$\overrightarrow{a} \times \overrightarrow{b} =(a_2 b_3 - a_3 b_2)\overrightarrow{i} + (a_3 b_1 - a_1 b_3)\overrightarrow{j} + (a_1 b_2 - a_2 b_1)\overrightarrow{k}$

with the goal of algebraically manipulating it into the form

$|\overrightarrow{a}|^2|\overrightarrow{b}|^2-(\overrightarrow{a}\cdot\overrightarrow{b})^2$

Consider the steps:

$|\overrightarrow{a} \times \overrightarrow{b}|^2$

$|\langle a_2 b_3 - a_3 b_2, a_3 b_1 - a_1 b_3, a_1 b_2 - a_2 b_1 \rangle|^2$

$\sqrt{(a_2 b_3 - a_3 b_2)^2 + (a_3 b_1 - a_1 b_3)^2 + (a_1 b_2 - a_2 b_1)^2}^2$

$(a_2 b_3 - a_3 b_2)^2 + (a_3 b_1 - a_1 b_3)^2 + (a_1 b_2 - a_2 b_1)^2$

which expands to:

$(a_3^2 b_2^2 - 2 a_2 a_3 b_3 b_2 + a_2^2 b_3^2) + (a_3^2 b_1^2 - 2 a_1 a_3 b_3 b_1 + a_1^2 b_3^2) + (a_2^2 b_1^2 - 2 a_1 a_2 b_2 b_1 + a_1^2 b_2^2)$

but I'm not really sure where to go from here. I could group the squared terms together:

$a_3^2 b_2^2 + a_2^2 b_1^2 + a_2^2 b_3^2 + a_3^2 b_1^2 + a_1^2 b_3^2 + a_1^2 b_2^2 - 2 a_2 a_3 b_3 b_2 - 2 a_1 a_3 b_3 b_1 - 2 a_1 a_2 b_2 b_1$

but no recognizable forms are really achieved (like the expanded expression for a dot product, $\overrightarrow{a} \cdot \overrightarrow{b} = a_1 b_1 + a_2 b_2 + a_3 b_3$).

How I algebraically prove this? Thanks for your help!

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  • $\begingroup$ Expanding everything is the simplest technology (and not too difficult). There is a bit more sophisticated way of proving this based on quaternion algebra. I don't know if that sheds enough more light to this to be worth your while (unless you want to learn more about quaternions also). You would still need to prove that the quaternion norm is multiplicative, and that is more or less equivalent to this equation. It does give room for some trickery. $\endgroup$ – Jyrki Lahtonen Sep 2 '15 at 17:57
  • $\begingroup$ If you find the number of terms daunting, then you can try the following. Everything in sight, the cross product, the inner product, the lengths, is rotation-invariant. Or, when viewed differently, independent of the choice of the right handed orthonormal basis. What this means is that it suffices to prove it in the case, where $\vec{a}$ is parallel to $x$-axis. Of course, you would then need to justify the said invariance. The more elegant ways of seeing that also require other machinery. If this is the first time you see this I really recommend that you just put your shoulder to the wheel. $\endgroup$ – Jyrki Lahtonen Sep 2 '15 at 18:05
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Having grouped your terms together as $$a_3^2 b_2^2 + a_2^2 b_1^2 + a_2^2 b_3^2 + a_3^2 b_1^2 + a_1^2 b_3^2 + a_1^2 b_2^2 - 2 a_2 a_3 b_3 b_2 - 2 a_1 a_3 b_3 b_1 - 2 a_1 a_2 b_2 b_1$$ we can add and subtract some extra terms (highlighted in colour below):- $$\begin{align}&a_3^2 b_2^2 + a_2^2 b_1^2 + a_2^2 b_3^2 + a_3^2 b_1^2 + a_1^2 b_3^2 + a_1^2 b_2^2 - 2 a_2 a_3 b_3 b_2 - 2 a_1 a_3 b_3 b_1 - 2 a_1 a_2 b_2 b_1\\&\color{red}{+a_1^2b_1^2+a_2^2b_2^2+a_3^2b_3^2\color{blue}{-a_1^2b_1^2-a_2^2b_2^2-a_3^2b_3^2}}\\&=(a_1^2+a_2^2+a_3^2)(b_1^2+b_2^2+b_3^2)-(a_1b_1+a_2b_2+a_3b_3)^2\\&=|\overrightarrow{a}|^2|\overrightarrow{b}|^2-(\overrightarrow{a}\cdot\overrightarrow{b})^2\end{align}$$

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Let me try.

You already have: $$\begin{eqnarray}LHS &=& a_1^2b_2^2 + a_1^2b_3^2 + a_2^2b_1^2 + a_2^2b_3^2 + a_3^2b_1^2 + a_3^2b_2^2 - 2 a_1a_2b_1b_2 - 2 a_2a_3b_2b_3 - 2a_3a_1b_3b_1\\ & =& a_1^2b_2^2 + a_1^2b_3^2 + a_2^2b_1^2 + a_2^2b_3^2 + a_3^2b_1^2 + a_3^2b_2^2 + a_1^2b_1^2 + a_2^2b_2^2 + a3^2b_3^2 -(a_1^2b_1^2 + a_2^2b_2^2 + a3^2b_3^2 + 2 a_1a_2b_1b_2 + 2 a_2a_3b_2b_3 + 2a_3a_1b_3b_1) \\ &=& (a_1^2+a_2^2+a_3^2)(b_1^2+b_2^2+b_3^2) - (a_1b_1+a_2b_2+a_3b_3)^2 \\ &=& RHS\end{eqnarray}$$

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