If $A = < 2, 6, - 3 >$ $A= <2 ,6 ,-3 >$ and $B = < 3, - 1, 5 >$ $B= <3 ,-1 ,5 >$ , what is $A \cdot B - | | A | | | | B | |$ $A*B -||A|| ||B||$ ?

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If $A = < 2, 6, - 3 >$ $A= <2 ,6 ,-3 >$ and $B = < 3, - 1, 5 >$ $B= <3 ,-1 ,5 >$ , what is $A \cdot B - | | A | | | | B | |$ $A*B -||A|| ||B||$ ?

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Answer 1 · 2016-01-12T17:12:29

$15 - \sqrt{1715}$ $15 - sqrt1715$

Explanation:

If $A$ $A$ and $B$ $B$ are vectors, then $A . B = 3 \sum i = 1 x_{a i} y_{b i}$ $A.B = sum_(i=1)^3 x_(ai)y_(bi)$ with $a_{i}, b_{i} \in {1, 2, 3}$ $a_i,b_i in {1,2,3}$ .

$A . B = 2 \cdot 3 + 6 \cdot (- 1) + 5 \cdot (- 3) = 6 - 6 - 15 = 15$ $A.B = 2*3 + 6*(-1) + 5*(-3) = 6 - 6 - 15 = 15$ .

$| | A | | = \sqrt{x_{a}^{2} + y_{a}^{2} + z_{a}^{2}}$ $||A|| = sqrt(x_a^2 + y_a^2 + z_a^2)$ , so $| | A | | = \sqrt{2^{2} + 6^{2} + {(- 3)}^{2}} = \sqrt{49}$ $||A|| = sqrt(2^2 + 6^2 + (-3)^2) = sqrt49$ and $| | B | | = \sqrt{3^{2} + {(- 1)}^{2} + 5^{2}} = \sqrt{35}$ $||B|| = sqrt(3^2 + (-1)^2 + 5^2) = sqrt(35)$

Hence $A . B - | | A | | \cdot | | B | | = 15 - \sqrt{35 \cdot 49} = 15 - \sqrt{1715}$ $A.B - ||A||*||B|| = 15 - sqrt(35*49) = 15 - sqrt(1715)$

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If $A = < 2, 6, - 3 >$ $A= <2 ,6 ,-3 >$ and $B = < 3, - 1, 5 >$ $B= <3 ,-1 ,5 >$ , what is $A \cdot B - | | A | | | | B | |$ $A*B -||A|| ||B||$ ?

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If $A = < 2, 6, - 3 >$ $A= <2 ,6 ,-3 >$ and $B = < 3, - 1, 5 >$ $B= <3 ,-1 ,5 >$ , what is $A \cdot B - | | A | | | | B | |$ $A*B -||A|| ||B||$ ?