What is the angle between $< 5, 7, 1 >$ $<5,7,1>$ and $< 5, 1, 7 >$ $<5,1,7>$ ?

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What is the angle between $< 5, 7, 1 >$ $<5,7,1>$ and $< 5, 1, 7 >$ $<5,1,7>$ ?

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Answer 1 · 2017-01-18T13:51:42

The angle is $58.7$ $58.7$ º

Explanation:

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

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

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

The dot product is

$\to A . \to B = ⟨ 5, 7, 1 ⟩ . ⟨ 5, 1, 7 ⟩ = 25 + 7 + 7 = 39$ $vecA.vecB=〈5,7,1〉.〈5,1,7〉=25+7+7=39$

The modulus of $\to A$ $vecA$ = $∥ ∥ ⟨ 5, 7, 1 ⟩ ∥ = \sqrt{25 + 49 + 1} = \sqrt{75}$ $∥〈5,7,1〉∥=sqrt(25+49+1)=sqrt75$

The modulus of $\to C$ $vecC$ = $∥ ∥ ⟨ 5, 1, 7 ⟩ ∥ = \sqrt{15 + 1 + 49} = \sqrt{75}$ $∥〈5,1,7〉∥=sqrt(15+1+49)=sqrt75$

So,

$cos θ = \frac{\to A . \to C}{∥ ∥ ∥ \to A ∥ \cdot ∥ \to C ∥ ∥ ∥} = \frac{39}{\sqrt{75} \cdot \sqrt{75}} = \frac{39}{75} = 0.52$ $costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=39/(sqrt75*sqrt75)=39/75=0.52$

$θ = 58.7$ $theta=58.7$ º

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What is the angle between $< 5, 7, 1 >$ $<5,7,1>$ and $< 5, 1, 7 >$ $<5,1,7>$ ?

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Explanation:

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