If $A = < 8, 3, - 1 >$ $A= <8 ,3 ,-1 >$ and $B = < 6, 4, 3 >$ $B= <6 ,4 ,3 >$ , what is $A \cdot B - | | A | | | | B | |$ $A*B -||A|| ||B||$ ?

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Answer 1 · 2018-07-01T12:24:54

Dot product is defined in $RR^3$ by

$(a_1,a_2,a_3)·(b_1,b_2,b_3)=a_1b_1+a_2b_2+a_3b_3$

By other hand vector module is defined by dot product by itself, it´s say $sqrt(veca·veca)$ in the same terms of above.

In our case

$vecA·vecB=8·6+3·4-1·3=57$

$abs(vecA)=sqrt(8^2+3^2+1^2)=sqrt74$

$abs(vecB)=sqrt(6^2+4^2+3^2)=sqrt61$

Finally: $vecA·vecB-abs(vecA)·abs(vecB)=57-sqrt74sqrt61=$

$=57-67.1863=-10.1863$

If A= <8 ,3 ,-1 >A=<8,3,−1>A= <8 ,3 ,-1 > and B= <6 ,4 ,3 >B=<6,4,3>B= <6 ,4 ,3 >, what is A*B -||A|| ||B||A⋅B−||A||||B||A*B -||A|| ||B||?