If #A= <-5 ,-4 ,-7 ># and #B= <2 ,-9 ,8 >#, what is #A*B -||A|| ||B||#?

1 Answer
Jun 21, 2018

#-13260 #

Explanation:

To answer this question we need three pieces:

  • #A*B#
  • #||A||#
  • #||B||#

but don't worry, they're all easy to compute!

  • #A*B# is the scalar product of two vectors. As the name suggests, it is a number (scalar), computed as follows: #A# and #B# must have the same length, and we must add the products of the coordinates of #A# and #B# in the same positions.

So, if #A=(a_1,a_2,...,a_n)# and #B=(b_1,b_2,...,b_n)#, we have

#A*B = a_1b_1+a_2b_2+...+a_nb_n = sum_{i=1}^n a_ib_i#

So, in your case, we have

#A*B = -5*2 + (-4)(-9) + (-7)*8 = -10+36-56=-30#

The other pieces, #||A||# and #||B||#, are called the norm of #A# and #B#. Actually, the norm is defined as the scalar product of a vector with itself: #||A|| = A*A#, we already know how to compute them:

#||A|| = (-5)(-5)+(-4)(-4)+(-7)(-7) = (-5)^2+(-4)^2+(-7)^2 = 25+16+49=90#

#||B|| = 2^2+(-9)^2+8^2 = 2+81+64=147#

We're finally ready!

#A*B - ||A|||B|| = -30 - 90*147 = -30-13230 = -13260#