Prove that $ a × b = - (b × a) $ for all $ a,b \in \mathbb{R}^{3} $.

Question

Problem: Prove that $ a × b = - (b × a) $ for all $ a,b \in \mathbb{R}^{3} $.

I am stuck with this question and unaware of what to do or where to start. If someone could help me, that would be greatly appreciated.

Answer 1

Let $\vec{a}=\langle a_1,a_2,a_3 \rangle$ and $\vec{b}=\langle b_1,b_2,b_3 \rangle$. Then,

$$\begin{align} \vec{a}\times\vec{b}:&=\langle a_2b_3-a_3b_2,a_3b_1-a_1b_3,a_1b_2-a_2b_1\rangle\\ &=-\langle a_3b_2-a_2b_3,a_1b_3-a_3b_1,a_2b_1-a_1b_2\rangle\\ &=-\langle b_2a_3-b_3a_2,b_3a_1-b_1a_3,b_1a_2-b_2a_1\rangle\\ &=:-\vec{b}\times\vec{a}. \end{align}$$

Answer 2

$a \times b$ is the unique vector $c$ such that: $$\forall x \in \mathbb R^3, \det(a,b,x)=\langle c,x \rangle $$ where $\det$ is the determinant in a direct othonrmal basis.

Since: $\det(b,a,x)=-\det(a,b,x)$ for all $x \in \mathbb R^3$, wecan say that the unique vector $c'$ such taht: $$\forall x \in \mathbb R^3, \det(b,a,x)= \langle c',x \rangle $$ is $-(a \times b)$ and conclude that $b \times a= - (a \times b) $