# How do you find the cross product a × b given a = ‹t,t^2,t^3›, b = ‹1, 6t, 8t^2› ?

$a X b = < 2 {t}^{4} , - 7 {t}^{3} , 5 {t}^{2} >$
If$a = < {a}_{1.} {a}_{2} , {a}_{3} > \mathmr{and} b = < {b}_{1} , {b}_{2} , {b}_{3} > , a X b = < {a}_{2} {b}_{3} - {a}_{3} {b}_{2} , {a}_{3} {b}_{1} - {a}_{1} {b}_{3} , {a}_{1} {b}_{2} - {a}_{2} {b}_{1} >$.
Here, $a X b = < 8 {t}^{4} - 6 {t}^{4} , {t}^{3} - 8 {t}^{3} , 6 {t}^{2} - {t}^{2} \ge < 2 {t}^{4} , - 7 {t}^{3} , 5 {t}^{2} >$.