If #A= <5 ,1 ,-2 ># and #B= <4 ,-2 ,3 >#, what is #A*B -||A|| ||B||#?

1 Answer
May 12, 2017

#12-sqrt(870)#

Explanation:

For two vectors #A# and #B# of the form #A=< A_x,A_y,A_z ># and #B=< B_x,B_y,B_z >,# the dot product #A*B# is given by #(a_x*b_x)+(a_y*b_y)+(a_z*b_z)#.

For #A=< 5,1,-2 ># and #B=< 4,-2,3 >#, we have:

#A*B=(5*4)+(1*-2)+(-2*3)=20-2-6=color(blue)12#

The next part is the product of the magnitudes of vectors #A# and #B#. The magnitude of a vector #A=< A_x,A_y,A_z ># is given by:

#|A|=sqrt((A_x)^2+(A_y)^2+(A_z)^2 )#

For the given vectors #A# and #B#:

#|A|=sqrt((5)^2+(1)^2+(-2)^2)=sqrt(25+1+4)=sqrt(30)#

#|B|=sqrt((4)^2+(-2)^2+(3)^2)=sqrt(16+4+9)=sqrt(29)#

We then have #|A|*|B|=sqrt(30)*sqrt(29)=sqrt(870)#

Then #A*B-|A||B|=12-sqrt(870)#.