If $A = < - 5, - 4, - 1 >$ $A= <-5 ,-4 ,-1 >$ and $B = < 2, - 9, 8 >$ $B= <2 ,-9 ,8 >$ , what is $A \cdot B - | | A | | | | B | |$ $A*B -||A|| ||B||$ ?

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Answer 1 · 2016-10-13T21:00:44

$¯ ¯ ¯ A \cdot ¯ ¯ ¯ B - ∣ ∣ ¯ ¯ ¯ A ∣ ∣ ∣ ∣ ¯ ¯ ¯ B ∣ ∣ \approx - 61.1$ $barA*barB - |barA||barB| ~~ -61.1$

For the dot-product of two vectors, one multiplies the corresponding components:

$¯ ¯ ¯ A \cdot ¯ ¯ ¯ B = (- 5) (2) + (- 4) (- 9) + (- 1) (8)$ $barA*barB = (-5)(2) + (-4)(-9) + (-1)(8)$

$¯ ¯ ¯ A \cdot ¯ ¯ ¯ B = 18$ $barA*barB = 18$

The magnitude of a vector is the square root of the sum of the squares of the components:

$∣ ∣ ¯ ¯ ¯ A ∣ ∣ = \sqrt{{(- 5)}^{2} + {(- 4)}^{2} + {(- 1)}^{2}}$ $|barA| = sqrt((-5)^2 + (-4)^2 + (-1)^2)$

$∣ ∣ ¯ ¯ ¯ A ∣ ∣ = \sqrt{42}$ $|barA| = sqrt(42)$

$∣ ∣ ¯ ¯ ¯ B ∣ ∣ = \sqrt{(2^{2} + {(- 9)}^{2} + {(8)}^{2})}$ $|barB| = sqrt((2^2 + (-9)^2 + (8)^2)$

$∣ ∣ ¯ ¯ ¯ B ∣ ∣ = \sqrt{149}$ $|barB| = sqrt(149)$

$¯ ¯ ¯ A \cdot ¯ ¯ ¯ B - ∣ ∣ ¯ ¯ ¯ A ∣ ∣ ∣ ∣ ¯ ¯ ¯ B ∣ ∣ = 18 - \sqrt{42} \sqrt{149}$ $barA*barB - |barA||barB| = 18 - sqrt(42)sqrt(149)$

$¯ ¯ ¯ A \cdot ¯ ¯ ¯ B - ∣ ∣ ¯ ¯ ¯ A ∣ ∣ ∣ ∣ ¯ ¯ ¯ B ∣ ∣ \approx - 61.1$ $barA*barB - |barA||barB| ~~ -61.1$

If A=<−5,−4,−1>A= <-5 ,-4 ,-1 > and B=<2,−9,8>B= <2 ,-9 ,8 >, what is A⋅B−||A||||B||A*B -||A|| ||B||?