How to use Cross Product Properites to do proof How do I proceed with a proof for this question?
Prove that:
\begin{equation}
       (a \times  b) \cdot (c \times d)  =  \begin{vmatrix}
    a \cdot c & b \cdot c \\
    a \cdot d & b \cdot d \\
    \end{vmatrix} 
\end{equation}
I have to use the cross product properties to do the proof:

So far I have taken the det of the right side:
\begin{equation}
(a \times b) \cdot (c \times d) = (a \cdot c) \cdot (b\cdot d) - (b \cdot c)(a \cdot d) 
\end{equation}
I don't understand how to use the properities after this step.
 A: Property 5 states that:
$$\color{blue}{u}\cdot (\color{red}{v}\times\color{green}{w})=(\color{blue}{u}\times\color{red}{v})\cdot\color{green}{w}$$
Applying property 5 to the L.H.S. we get:
$\color{blue}{(a\times b)}\cdot (\color{red}{c}\times \color{green}{d}) = (\color{blue}{(a\times b)}\times \color{red}{c})\cdot \color{green}{d}$
Property 1 states that:
$$\color{blue}{u}\times \color{red}{v}=\color{red}{-v}\times\color{blue}{u}$$
So, applying property 1 to our previous step:
$(\color{blue}{(a\times b)}\times \color{red}{c})\cdot \color{green}{d}=(\color{red}{-c}\times \color{blue}{(a\times b)})\cdot \color{green}{d}$
Now that we have something of the form $u\times (v\times w)$, we can use property 6.
Continue the problem using property 6 to get rid of the $\times$ products in the expression and replace it with dot products, addition, and subtraction.  Continue simplifying to get to the result.

Property 6 states that:
$$\color{blue}{u}\times(\color{red}{v}\times\color{green}{w})=(\color{blue}{u}\cdot\color{green}{w})\color{red}{v}-(\color{blue}{u}\cdot\color{red}{v})\color{green}{w}$$
Using prop6 should only affect what is inside of the black parenthesis that I have written previously.  
$=(\color{blue}{-c}\times (\color{red}{a}\times\color{green}{b}))\cdot d = ((\color{blue}{-c}\cdot\color{green}{b})\color{red}{a}-(\color{blue}{-c}\cdot\color{red}{a})\color{green}{b})\cdot d$
A: Using property 5, we get $(a \times  b) \cdot (c \times d) = ((a \times  b) \times c) \cdot d$. Then use property 1 to get $((a \times  b) \times c) \cdot d = (-c \times (a \times  b)) \cdot d$. Now we can apply property 6 to get $(-c \times (a \times  b)) \cdot d = -((c\cdot b)a-(c\cdot a)b)\cdot d$. If you simplify this expression using linearity of the dot product, you should get your expansion for the determinant.
A: $$\begin{align}
(\vec a\times \vec b)\cdot(\vec c\times \vec d)&=((\vec a\times \vec b)\times \vec c)\cdot \vec d \tag 1 \\\\
&=(-\vec c\times (\vec a\times \vec b))\cdot \vec d \tag 2 \\\\
&=((\vec c\cdot \vec a)\vec b-(\vec c\cdot \vec b)\vec a)\cdot \vec d\tag 3\\\\
&=(\vec c\cdot \vec a)(\vec b\cdot \vec d)-(\vec c\cdot \vec b)(\vec a\cdot \vec d)
\end{align}$$
To arrive at $(1)$, we used Property $5$
To go from $(1)$ to $(2)$, we used Property $1$
To go from $(2)$ to $(3)$, we used Property $6$
A: from the 5th property we have: $(a\times b).(c\times d)=((a\times b)\times c).d$
then from the 1th property: $((a\times b)\times c).d=-(c\times (a\times b)).d$
then from the 6th property:$-(c\times (a\times b)).d=-((c.b)a-(c.a)b).d$
so:
$$\begin{align}(a\times b).(c\times d)&=((a\times b)\times c).d\\
                                      &=-(c\times (a\times b)).d\\
                                      &=-((c.b)a-(c.a)b).d\\
                                      &=-((c.b)(a.d)-(c.a)(b.d))\\                         &=((c.a)(b.d)-(c.b)(a.d))\\                         &=(a.c)(b.d)-(b.c)(a.d)\\                          &=\begin{vmatrix}a.c & b.c\\                                                  a.d & b.d\end{vmatrix}\end{align}$$
