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

1 Answer
Feb 16, 2018

#color(blue)(-29.46)color(white)(888)# (2 .d.p)

Explanation:

#bbA=[(3),(1),(-4)]# , #color(white)(88)bbB=[(4),(-2),(3)]#

#bb(A*B)#

This is the Dot Product and is defined as:

#bb(A*B)=||bbA||*||bbB||*costheta#

The multiplication of #bb(A*B)# is different from the normal way in which we multiply in algebra. Normally we multiply in the following way:

#(a+b+c)(d+e+f)=ad+ae+af+bd+be+bf+cd+ce+cf#

In the dot product we multiply as follows:

#(a+b+c)(d+e+f)=ad+be+cf#

So we are just multiplying corresponding components and summing the results. This is where the alternative name Inner Product comes from.

For some vector #bbA=[(a),(b),(c)]#

#||bbA||# is the magnitude of the vector #bbA# and is defined as:

#||bbA||=sqrt(a^2+b^2+c^2)#

This is just the distance formula found in coordinate geometry.

To the example:

First we find the dot product of #bbA# and #bbB#

#bb(A*B)=(3xx4)+(1xx-2)+(-4xx3)=-2#

Now we find the magnitudes of #bbA# and #bbB#

#||bbA||=sqrt((3)^2+(1)^2+(-4)^2)=sqrt(9+1+16)=sqrt(26)#

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

Now we require:

#bb(A*B)-||bbA||||bbB||#

#-2-sqrt(26)sqrt(29)=color(blue)(-29.46)color(white)(888)# (2 .d.p)