Calculus: Early Transcendentals 8th Edition

$(a \times b) \cdot (c \times d)=\begin{vmatrix} a \cdot c&b \cdot c\\a \cdot d& b \cdot d\end{vmatrix}$
Let $a= \lt a_1,a_2,a_3 \gt$; $b=\lt b_1,b_2,b_3 \gt$ , $c=\lt c_1,c_2,c_3 \gt$ and $d=\lt d_1,d_2,d_3 \gt$ $a \times b=\begin{vmatrix} i&j&k\\a_1&a_2&a_3 \\b_1&b_2&b_3 \end{vmatrix}$ Using properties of determinants, we can write $a \times b= \lt a_2b_3-a_3b_2,b_1a_3-b_3a_1,a_1b_2-a_2b_1 \gt$ $c \times d=\begin{vmatrix} i&j&k\\c_1&c_2&c_3 \\d_1&d_2&d_3 \end{vmatrix}$ Using properties of determinants, we can write $c \times d= \lt c_2d_3-c_3d_2,d_1c_3-d_3c_1,c_1d_2-c_2d_1 \gt$ Now dot product: $(a \times b) \cdot (c \times d)=(a_2b_3-a_3b_2)(c_2d_3-c_3d_2)+(b_1a_3-b_3a_1)(d_1c_3-d_3c_1)+(a_1b_2-a_2b_1)(c_1d_2-c_2d_1 )$ Write the determinant of the right side of the equation in terms of components. $\begin{vmatrix} a \cdot c&b \cdot c\\a \cdot d& b \cdot d\end{vmatrix} =(a \cdot c) (b.d)-(b.c)(a.d)$ $=(a_1c_1+a_2c_2+a_3c_3)(b_1d_1+b_2d_2+b_3d_3)-(b_1c_1+b_2c_2+b_3c_3)(a_1d_1+a_2d_2+a_3d_3)$ $=a_1b_2c_1d_2+a_1b_3c_1d_3+a_2b_1c_2d_1+a_2b_3c_2d_3+a_3c_3b_1d_1+a_3c_3b_2d_2-a_2b_1c_1d_2-a_3b_1c_1d_3-a_1b_2c_2d_1-a_3b_2c_2d_3-a_1b_3c_3d_1-a_2b_3c_3d_2$ Hence, $(a \times b) \cdot (c \times d)=\begin{vmatrix} a \cdot c&b \cdot c\\a \cdot d& b \cdot d\end{vmatrix}$