Example 2.1.2: Dot and cross products

With (2.1.5),

Expanding the determinant (2.1.8) along its first row,

that is $a_x(-2-12)-a_y(4-6)+a_z(12+3)$, and

The cross product must be perpendicular to both factors: $A\cdot (A\times B)=-28-2+30=0$ and $B\cdot (A\times B)=-42+12+30=0$. With $A=3$ and $B=7$, the angle between the vectors is ${\theta }_{AB}=\arccos (4/21)\approx 79.0^{\circ }$. The vectors are not far from perpendicular, which is why the dot product is small compared with $AB=21$. Lagrange's identity gives $|A\times B{|}^2=A^2{B}^2-(A\cdot B{)}^2=441-16=425$, in agreement with $(-14{)}^2+2^2+15^2=425$.