If #A= <6 ,9 ,-1 ># and #B= <-4 ,-1 ,4 >#, what is #A*B -||A|| ||B||#?

1 Answer
Trevor Ryan.
Jan 7, 2016

#62.401#.

Explanation:

Definitions : Let #A=(a_1,a_2,....,a_n) and B=(b_1,b_2,....,b_n)# be any 2 vectors in a real or complex finite dimensional vector space X. Then we define:

  1. The Euclidean inner product (dot product) of A and B as the real or complex number given by #A*B= a_1b_1+a_2b_2+......+a_nb_n#.
  2. The norm of A as the real or complex number given by #||A||=sqrt(a_1^2+a_2^2+......+a_n^2)#.

Applying these 2 definitions to the given 3 dimensional vectors we get :

#A*B=(6,9,-1)*(-4,-1,4)#

#=(6xx-4)+(9xx-1)+(-1xx4)#

#=-24-9-4#

#=-37#.

#||A||=|| (6,9,-1) || = sqrt(6^2+9^2+1^2)=sqrt118#.

Similarly #||B||= sqrt33#.

#therefore A*B-||A|| ||B|| = -37-(sqrt118)(sqrt33)#

#=-37-sqrt(118xx33)#

#=62.401#.

Related questions
See all questions in Vectors
Impact of this question
1605 views around the world
You can reuse this answer
Creative Commons License