Find the vector, not with determinants, but by using properties of cross products. (j – k) x (k – i)
2 Answers

use the distributive properties of the cross product:
(j – k) x (k – i) = (jk) x k – (jk) x i =
j x k – k x k – ( j x i – k x i ) = j x k – k x k – j x i + k x i
Now k x k = 0
The other 3 you can do by visualizing it geometrically: draw them out and use the right hand rule for the direction. The length of the vector that results in each of these three must = 1 since the norm of a x b is absinΘ, and in all these 3 cases we have i=j=k = 1 and Θ= pi/2 so sin Θ = 1. :
j x k = i
j x i = – k
k x i = j
So we get i – (k) + j = i + j + k

Cross Products Property