What is the angle between $< 2, - 3, - 1 >$ $<2,-3,-1 >$ and $< - 8, 3, - 5 >$ $< -8,3,-5>$ ?

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What is the angle between $< 2, - 3, - 1 >$ $<2,-3,-1 >$ and $< - 8, 3, - 5 >$ $< -8,3,-5>$ ?

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Answer 1 · 2016-02-29T17:40:29

$cos θ = \frac{< 2, - 3, - 1 > \cdot < - 8, 3, - 5 >}{(\sqrt{2^{2} + (- 3^{2}) + {(- 1)}^{2}}) (\sqrt{{(- 8)}^{2} + 3^{2} + {(- 5)}^{2}})}$ $cos theta=(<2,-3,-1>*<-8,3,-5>)/((sqrt(2^2+(-3^2)+(-1)^2)) (sqrt((-8)^ 2+3^2+(-5)^2))$
= $\frac{2 (- 8) + (- 3) (3) + (- 1) (- 5)}{(\sqrt{14} \sqrt{98})}$ $(2(-8)+(-3)(3)+(-1)(-5))/((sqrt(14)sqrt(98))$
= $- \frac{20}{\sqrt{1372}}$ $-20/sqrt1372$
$θ = {cos}^{- 1} (- \frac{20}{\sqrt{1372}}) = {122.68}^{\circ}$ $theta=cos^-1(-20/sqrt(1372))=122.68^@$

Explanation:

To find theta take the inverse cosine of the dot product of the given vectors divided by the product of the magnitude of each of the vectors.

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What is the angle between $< 2, - 3, - 1 >$ $<2,-3,-1 >$ and $< - 8, 3, - 5 >$ $< -8,3,-5>$ ?

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Explanation:

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