What is the angle between $< - 4, 8, 0 >$ $<-4,8,0>$ and $< - 3, 6, - 2 >$ $< -3,6,-2>$ ?

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What is the angle between $< - 4, 8, 0 >$ $<-4,8,0>$ and $< - 3, 6, - 2 >$ $< -3,6,-2>$ ?

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Answer 1 · 2016-10-16T18:04:04

$(\to u, \to v) = 0.29 r a d$ $(vecu,vecv)=0.29 rad$

Explanation:

Let $\to u = (- 4, 8, 0)$ $vecu=(-4,8,0)$
and $\to v = (- 3, 6 - 2)$ $vecv=(-3,6-2)$
Use the dot product definition, $\to u \cdot \to v = ∣ \to u ∣ \cdot ∣ \to v ∣ \cdot cos (\to u, \to v)$ $vecu*vecv=∣vecu∣*∣vecv∣*cos(vecu,vecv)$
$\to u . \to v = 12 + 48 + 0 = 60$ $vecu.vecv=12+48+0=60$
$∣ ∣ ∣ \to u ∣ ∣ = \sqrt{16 + 64 + 0} = \sqrt{80}$ $∣∣vecu∣∣= sqrt(16+64+0)=sqrt80$
$∣ ∣ ∣ \to v ∣ ∣ = \sqrt{9 + 36 + 4} = \sqrt{49} = 7$ $∣∣vecv∣∣=sqrt(9+36+4)=sqrt49=7$
So $cos (\to u, \to v) = \frac{\to u . \to v}{∣ ∣ \to u ∣ ∣ \to v ∣ ∣} = \frac{60}{\sqrt{80} \cdot 7}$ $cos(vecu,vecv)=(vecu.vecv)/(∣vecu∣∣vecv∣)=60/(sqrt80*7)$
$(\to u, \to v) = arccos (\frac{60}{7 \sqrt{80}}) = 0.29 r a d$ $(vecu,vecv)=arccos(60/(7sqrt(80)) )=0.29rad$

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What is the angle between $< - 4, 8, 0 >$ $<-4,8,0>$ and $< - 3, 6, - 2 >$ $< -3,6,-2>$ ?

1 Answer

Explanation:

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What is the angle between <−4,8,0><-4,8,0> and <−3,6,−2>< -3,6,-2>?

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What is the angle between $< - 4, 8, 0 >$ $<-4,8,0>$ and $< - 3, 6, - 2 >$ $< -3,6,-2>$ ?