# Chapter 12 - Vectors and the Geometry of Space - 12.4 The Cross Product - 12.4 Exercises - Page 862: 51

$a \times (b \times c)+b \times (c \times a)+c \times (a \times b)=0$

#### Work Step by Step

Let $a= a_1i+a_2j+a_3k$; $b=b_1i+b_2j+b_3k$ and $c=c_1i+c_2j+c_3k$ $b \times c=\begin{vmatrix} i&j&k\\b_1&b_2&b_3 \\c_1&c_2&c_3\end{vmatrix}$ Using properties of determinants, we can write $b \times c= i(b_2c_3-c_2b_3)-j(b_1c_3-b_3c_1)+k(b_1c_2-b_2c_1)$ $a \times (b \times c)=\begin{vmatrix} i&j&k\\a_1&a_2&a_3\\b_2c_3-c_2b_3&-(b_1c_3-b_3c_1)&b_1c_2-b_2c_1\end{vmatrix}$ Using properties of determinants, we can write $a \times (b \times c)= \lt (a_2(b_1c_2-b_2c_1)-a_3(b_3c_1-b_1c_3)),(a_3(b_2c_3-b_3c_2)-a_1(b_1c_2-b_2c_1)),(a_1(b_3c_1-b_1c_3)-a_2(b_2c_3-b_3c_2))\gt$ $a \times (b \times c)= \lt a_2b_1c_2-a_2b_2c_1-a_3b_3c_1+a_3b_1c_3,a_3b_2c_3-a_3b_3c_2-a_1b_1c_2+a_1b_2c_1,a_1b_3c_1-a_1b_1c_3-a_2b_2c_3+a_2b_3c_2\gt$ $a \times (b \times c)= \lt b_1(a \cdot c)-c_1(a \cdot b),b_2(a \cdot c)-c_2(a \cdot b), b_3(a \cdot c)-c_3(a \cdot b)\gt$ $a \times (b \times c)= \lt b_1(a \cdot c) ,b_2(a \cdot c),b_3(a \cdot c) \gt - \lt c_1(a \cdot b),c_2(a \cdot b), c_3(a \cdot b)\gt$ $a \times (b \times c)= (a \cdot c)\lt b_1 ,b_2,b_3\gt - (a \cdot b)\lt c_1,c_2, c_3\gt$ $a \times (b \times c)= (a \cdot c)b - (a \cdot b) c$ $a \times (b \times c)+b \times (c \times a)+c \times (a \times b)= (a \cdot c)b - (a \cdot b) c+(b \cdot a)c - (b \cdot c) a+(c \cdot b)a - (c \cdot a) b$ $a \times (b \times c)+b \times (c \times a)+c \times (a \times b)= 0+0+0=0$ Hence, $a \times (b \times c)+b \times (c \times a)+c \times (a \times b)=0$

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