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

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Answer 1 · 2016-03-28T09:01:26

$A \cdot B - | A | \cdot | B | = - 17 - 92.4124 = - 109.4124$ $A*B-|A|*|B|=-17-92.4124=-109.4124$

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

$A \cdot B = (a_{1} \cdot b_{1}) + (a_{2} \cdot b_{2}) + (a_{3} \cdot b_{3})$ $A*B=(a_1*b_1)+(a_2*b_2)+(a_3*b_3)$ .

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

$A \cdot B = (6 \cdot 5) + (- 5) \cdot 4 + (3 \cdot (- 9)$ $A*B=(6*5)+(-5)* 4+(3*(-9)$ or

$A \cdot B = 30 - 20 - 27 = - 17$ $A*B=30-20-27=-17$

$| A |$ $|A|$ represents the magnitude of vector $A$ $A$ and is given by $\sqrt{a_{1}^{2} + a_{2}^{2} + a_{3}^{2}}$ $sqrt(a_1^2+a_2^2+a_3^2)$ .

Hence $| A | \cdot | B |$ $|A|*|B|$

= $(\sqrt{6^{2} + {(- 5)}^{2} + 3^{2}}) \cdot (\sqrt{5^{2} + 4^{2} + {(- 9)}^{2}})$ $(sqrt(6^2+(-5)^2+3^2)) * (sqrt(5^2+4^2+(-9)^2))$

= $(\sqrt{36 + 25 + 9}) \cdot (\sqrt{25 + 16 + 81})$ $(sqrt(36+25+9)) * (sqrt(25+16+81))$

= $\sqrt{70} \cdot \sqrt{122} = 8.3666 \cdot 11.0454 = 92.4124$ $sqrt70*sqrt122=8.3666*11.0454=92.4124$

Hence $A \cdot B - | A | \cdot | B | = - 17 - 92.4124 = - 109.4124$ $A*B-|A|*|B|=-17-92.4124=-109.4124$

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