Chapter 8 - Polar Coordinates; Vectors - Section 8.7 The Cross Product - 8.7 Assess Your Understanding - Page 652: 45

$\sqrt {166}$

1. Find the vectors $\vec {P_1P_2}=< 1, 2, 3>$ and $\vec {P_1P_3}=<-2,3,0>$ 2. Find the cross product $\vec {P_1P_2}\times\vec {P_1P_3} = <(2(0)-3(3)),(3(-2)-1(0)),(1(3)-2(-2))>=<-9,-6,7>$ 3. We have the area $A=||\vec {P_1P_2}\times\vec {P_1P_3} || =\sqrt {9^2+6^2+7^2}=\sqrt {166}$