How to find the resulting vector by performing the algebraic multiplication (not determinant) of the elements?: (j-k) * (K-i)

#(j-k)*(K-i)#

1 Answer
Somebody N.
Apr 3, 2018

#bbi+bbj+bbk#

Explanation:

Vectors:

#j=bbj#

#k=bbk#

#i=bbi#

#(j-k)=bbj-bbk#

#(k-i)=-bbi+bbk#

#bbj-bbkxx-bbi+bbk=bbj*-bbi+bbj*bbk-bbk*-bbi-bbk*bbk#

The products of the unit vectors is best looked at as cyclic permutations.

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Looking at the diagram. If we multiply #bbixxbbj# we move in a clockwise direction to the next unit vector #bbk#. If we multiply #bbjxxbbi# we move in a anti-clockwise direction to the next unit vector #-bbk#. So using this idea,( you need to pay attention to the signs of the unit vectors).

#bbj*-bbi+bbj*bbk-bbk*-bbi-bbk*bbk#

#bbj*-bbi=bbk#

#bbj*bbk=bbi#

#-bbk*-bbi=-(bbk*bbi)=-(-bbj)=bbj#

#-bbk*bbk=0#*

This is because by the vector product definition:

#axxb=||a||*||b||sin(theta)*n#

Angle between #-bbk# and #bbk# is #theta = 0#, #sin(0)=0#

So we have:

#(j-k)(k-i)=bbi+bbj+bbk#

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