If $A= <2 ,6 ,-3 >$ and $B= <-8 ,2 ,7 >$ , what is $A*B -||A|| ||B||$ ?

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Answer 1 · 2017-11-25T07:24:55

The answer is $=-100.7$

The vectors are

$vecA= <2,6,-3>$

$vecB = <-8,2,7>$

The modulus of $vecA$ is $=||vecA||=||<2,6,-3>||=sqrt(2^2+(6)^2+(-3)^2)=sqrt(4+36+9)=sqrt49=7$

The modulus of $vecB$ is $=||vecB||=||<-8,2,7>||=sqrt((-8)^2+^2(2)^2+)=sqrt(64+4+49)=sqrt117$

Therefore,

$||vecA|| xx||vecB||=7*sqrt117=7sqrt117$

The dot product is

$vecA.vecB= <2,6,-3> .<-8,2,7> =(2xx-8)+(6xx2)+(-3xx7)=-16+12-21=-25$

Therefore,

$vecA.vecB-||vecA|| xx||vecB||=-25-7sqrt117= -100.7$

If A= <2 ,6 ,-3 >A= <2 ,6 ,-3 > and B= <-8 ,2 ,7 >B= <-8 ,2 ,7 >, what is A*B -||A|| ||B||A*B -||A|| ||B||?