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

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

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Answer 1 · 2016-10-19T09:56:51

Angle between $< - 8, 2, 8 >$ $< -8,2,8>$ and $< 2, - 3, 5 >$ $< 2,-3,5>$ is ${75.28}^{o}$ $75.28^o$

Explanation:

Angle between two vectors $\to u = a_{1} ˆ i + b_{1} ˆ j + c_{1} ˆ k$ $vecu=a_1hati+b_1hatj+c_1hatk$ or $< a_{1}, b_{1}, c_{1} >$ $< a_1,b_1,c_1>$

and $\to v = a_{2} ˆ i + b_{2} ˆ j + c_{2} ˆ k$ $vecv=a_2hati+b_2hatj+c_2hatk$ or $< a_{2}, b_{2}, c_{2} >$ $< a_2,b_2,c_2>$ is given by

$cos θ = \frac{(\to u \cdot \to v)}{(∣ ∣ \to u ∣ ∣ \cdot ∣ ∣ \to v ∣ ∣)}$ $costheta=((vecu*vecv))/((|vecu|*|vecv|))$ ,

where $\to u \cdot \to v = a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2}$ $vecu*vecv=a_1a_2+b_1b_2+c_1c_2$

and $∣ ∣ \to u ∣ ∣$ $|vecu|$ or $∣ ∣ \to v ∣ ∣$ $|vecv|$ are magnitudes of vectors $\to u$ $vecu$ or $\to v$ $vecv$ and here they are

$\sqrt{a_{1}^{2} + b_{1}^{2} + c_{1}^{2}}$ $sqrt(a_1^2+b_1^2+c_1^2)$ and $\sqrt{a_{2}^{2} + b_{2}^{2} + c_{2}^{2}}$ $sqrt(a_2^2+b_2^2+c_2^2)$

Hence angle between $< - 8, 2, 8 >$ $< -8,2,8>$ and $< 2, - 3, 5 >$ $< 2,-3,5>$ is given by

$cos θ = \frac{(- 8) \times 2 + 2 \times (- 3) + 8 \times 5}{\sqrt{{(- 8)}^{2} + 2^{2} + 8^{2}} \times \sqrt{{(2)}^{2} + {(- 3)}^{2} + 5^{2}}}$ $costheta=((-8)xx2+2xx(-3)+8xx5)/(sqrt((-8)^2+2^2+8^2)xxsqrt((2)^2+(-3)^2+5^2))$

= $\frac{- 16 - 6 + 40}{\sqrt{64 + 4 + 64} \times \sqrt{4 + 9 + 25}}$ $(-16-6+40)/(sqrt(64+4+64)xxsqrt(4+9+25))$

= $\frac{18}{\sqrt{132} \times \sqrt{38}}$ $18/(sqrt132xxsqrt38)$

= $\frac{18}{11.4891 \times 6.1644}$ $18/(11.4891xx6.1644)$

= $0.25415$ $0.25415$

and $θ = {75.28}^{o}$ $theta=75.28^o$

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

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Explanation:

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