Can't we factor out a constant in the cross product?

Question

I have the vectors $A=a\hat e_x$ and $B=a\hat e_y$, so $$ A\times B = \begin{vmatrix} \hat e_x &\hat e_y & \hat e_z \\ a & 0 & 0\\ 0 & a &0 \end{vmatrix}=\hat za^2 $$ Q1: But why is the following wrong \begin{align} A\times B &=a\hat e_x\times a\hat e_y\\ &=(a\hat e_x)\times (a\hat e_y) \tag 1\\ &=(a\hat e_x)\times (\hat e_ya) \tag 2\\ &=a(\hat e_x)\times (\hat e_y)a \tag 3\\ &=a((\hat e_x)\times (\hat e_y)) \tag 4\\ &=a(\hat e_x\times \hat e_y) \tag 5\\ &=a\hat e_z \quad ? \end{align}

Q2: Also, why is this wrong \begin{align} A\times B &= \begin{vmatrix} \hat e_x &\hat e_y & \hat e_z \\ a & 0 & 0\\ 0 & a &0 \end{vmatrix} \\ &= a \begin{vmatrix} \hat e_x &\hat e_y & \hat e_z \\ 1 & 0 & 0\\ \tag 6 0 & 1 &0 \end{vmatrix}\\ &=a\hat z \quad ? \end{align}

Answer 1

$$=a(\hat e_x)\times (\hat e_y)a \tag 3$$ $$=a((\hat e_x)\times (\hat e_y)) \tag 4$$

What happens here...?

You may be mixing it up with the distributive property of the cross product over addition.

Q2: Also, why is this wrong \begin{align} A\times B &= \begin{vmatrix} \hat e_x &\hat e_y & \hat e_z \\ a & 0 & 0\\ 0 & a &0 \end{vmatrix} \\ &= a \begin{vmatrix} \hat e_x &\hat e_y & \hat e_z \\ 1 & 0 & 0\\ \tag 6 0 & 1 &0 \end{vmatrix}\\ &=a\hat z \quad ? \end{align}

For starters, the determinant notation for computing the cross product is actually just a mnemonic. But if you want to use properties of determinants, you should note that the determinant is linear in its columns and rows, so for example: $$\begin{vmatrix} a & b & c \\ \color{purple}{k}d& \color{purple}{k}e & \color{purple}{k}f \\ g & h & i \end{vmatrix}=\color{purple}{k}\begin{vmatrix} a & b & c \\ d& e & f \\ g & h & i \end{vmatrix}$$ This means that if you want to factor out the $a$, you do that for the second and the third row: $$\begin{vmatrix} \hat e_x &\hat e_y & \hat e_z \\ \color{blue}{a}& 0 & 0\\ 0 & \color{red}{a} &0 \end{vmatrix}=\color{blue}{a}\color{red}{a}\begin{vmatrix} \hat e_x &\hat e_y & \hat e_z \\ 1 & 0 & 0\\ 0 & 1 &0 \end{vmatrix}=a^2\begin{vmatrix} \hat e_x &\hat e_y & \hat e_z \\ 1 & 0 & 0\\ 0 & 1 &0 \end{vmatrix}$$

Answer 2

We have that by cross product definition

$$A\times B =a\hat e_x\times a\hat e_y=a^2( e_x\times \hat e_y)$$

and by the properties of the determinant

$$A\times B = \begin{vmatrix} \hat e_x &\hat e_y & \hat e_z \\ a & 0 & 0\\ 0 & a &0 \end{vmatrix} = a^2 \begin{vmatrix} \hat e_x &\hat e_y & \hat e_z \\ 1 & 0 & 0\\ \tag 6 0 & 1 &0 \end{vmatrix}$$

Answer 3

Your first argument is wrong because you lost an $a$ between lines (3) and (4).

Your second argument is wrong because line (6) should have $a^2$, not $a$. You can confirm this just by evaluating the determinant, as you did at the beginning of your post. You have somehow misunderstood the rules for removing common factors from determinants.