Trying to show $|\overrightarrow{a}\times\overrightarrow{b}|^2=|\overrightarrow{a}|^2|\overrightarrow{b}|^2-(\overrightarrow{a}⋅\overrightarrow{b})^2$

Question

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!

Answer 1

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):- $$\begin{align}&a_3^2 b_2^2 + a_2^2 b_1^2 + a_2^2 b_3^2 + a_3^2 b_1^2 + a_1^2 b_3^2 + a_1^2 b_2^2 - 2 a_2 a_3 b_3 b_2 - 2 a_1 a_3 b_3 b_1 - 2 a_1 a_2 b_2 b_1\\&\color{red}{+a_1^2b_1^2+a_2^2b_2^2+a_3^2b_3^2\color{blue}{-a_1^2b_1^2-a_2^2b_2^2-a_3^2b_3^2}}\\&=(a_1^2+a_2^2+a_3^2)(b_1^2+b_2^2+b_3^2)-(a_1b_1+a_2b_2+a_3b_3)^2\\&=|\overrightarrow{a}|^2|\overrightarrow{b}|^2-(\overrightarrow{a}\cdot\overrightarrow{b})^2\end{align}$$

Answer 2

Let me try.

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