If #A= <6 ,-5 ,3 ># and #B= <5 ,4 ,-9 >#, what is #A*B -||A|| ||B||#?

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Answer 1 · 2016-03-28T09:01:26

#A*B-|A|*|B|=-17-92.4124=-109.4124#

Where #A*B# represents the dot product between vectors #A# and #B#. This is given by:

#A*B=(a_1*b_1)+(a_2*b_2)+(a_3*b_3)#.

Hence for #A=<6,-5,3># and #B=<5,4,-9>#

#A*B=(6*5)+(-5)* 4+(3*(-9)# or

#A*B=30-20-27=-17#

#|A|# represents the magnitude of vector #A# and is given by #sqrt(a_1^2+a_2^2+a_3^2)#.

Hence #|A|*|B|#

= #(sqrt(6^2+(-5)^2+3^2)) * (sqrt(5^2+4^2+(-9)^2))#

= #(sqrt(36+25+9)) * (sqrt(25+16+81))#

= #sqrt70*sqrt122=8.3666*11.0454=92.4124#

Hence #A*B-|A|*|B|=-17-92.4124=-109.4124#