What is the angle between $< 8, 7, 6 >$ $<8,7,6>$ and $< - 1, 6, 1 >$ $<-1,6,1>$ ?

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What is the angle between $< 8, 7, 6 >$ $<8,7,6>$ and $< - 1, 6, 1 >$ $<-1,6,1>$ ?

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Answer 1 · 2017-01-18T09:33:28

The angle is $57.9$ $57.9$ º

Explanation:

The angle between $\to A$ $vecA$ and $\to B$ $vecB$ is given by the dot product definition.

$\to A . \to B = ∥ \to A ∥ \cdot ∥ \to B ∥ cos θ$ $vecA.vecB=∥vecA∥*∥vecB∥costheta$

Where $θ$ $theta$ is the angle between $\to A$ $vecA$ and $\to B$ $vecB$

The dot product is

$\to A . \to B = ⟨ 8, 7, 6 ⟩ . ⟨ - 1, 6, 1 ⟩ = - 8 + 42 + 6 = 40$ $vecA.vecB=〈8,7,6〉.〈-1,6,1〉=-8+42+6=40$

The modulus of $\to A$ $vecA$ = $∥ ∥ ⟨ 8, 7, 6 ⟩ ∥ = \sqrt{64 + 49 + 36} = \sqrt{149}$ $∥〈8,7,6〉∥=sqrt(64+49+36)=sqrt149$

The modulus of $\to B$ $vecB$ = $∥ ∥ ⟨ - 1, 6, 1 ⟩ ∥ = \sqrt{1 + 36 + 1} = \sqrt{38}$ $∥〈-1,6,1〉∥=sqrt(1+36+1)=sqrt38$

So,

$cos θ = \frac{\to A . \to B}{∥ ∥ ∥ \to A ∥ \cdot ∥ \to B ∥ ∥ ∥} = \frac{40}{\sqrt{38} \cdot \sqrt{149}} = 0.53$ $costheta=(vecA.vecB)/(∥vecA∥*∥vecB∥)=40/(sqrt38*sqrt149)=0.53$

$θ = 57.9$ $theta=57.9$ º

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What is the angle between $< 8, 7, 6 >$ $<8,7,6>$ and $< - 1, 6, 1 >$ $<-1,6,1>$ ?

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Explanation:

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