What is the angle between $< - 1, 0, 6 >$ $<-1,0,6 >$ and $< 2, - 1, 5 >$ $< 2,-1,5 >$ ?

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What is the angle between $< - 1, 0, 6 >$ $<-1,0,6 >$ and $< 2, - 1, 5 >$ $< 2,-1,5 >$ ?

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Answer 1 · 2018-04-19T06:55:04

The angle is $= {32.8}^{\circ}$ $=32.8^@$

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

$\to A . \to B = ⟨ - 1, 0, 6 ⟩ . ⟨ 2, - 1, 5 ⟩ = - 2 + 0 + 30 = 28$ $vecA.vecB=〈-1,0,6〉.〈2,-1,5〉=-2+0+30=28$

The modulus of $\to A$ $vecA$ = $∥ ∥ ⟨ - 1, 0, 6 ⟩ ∥ = \sqrt{1 + 0 + 36} = \sqrt{37}$ $∥〈-1,0,6〉∥=sqrt(1+0+36)=sqrt37$

The modulus of $\to B$ $vecB$ = $∥ ∥ ⟨ 2, - 1, 5 ⟩ ∥ = \sqrt{4 + 1 + 25} = \sqrt{30}$ $∥〈2,-1,5〉∥=sqrt(4+1+25)=sqrt30$

So,

$cos θ = \frac{\to A . \to B}{∥ ∥ ∥ \to A ∥ \cdot ∥ \to B ∥ ∥ ∥} = \frac{28}{\sqrt{37} \cdot \sqrt{30}} = 0.84$ $costheta=(vecA.vecB)/(∥vecA∥*∥vecB∥)=28/(sqrt37*sqrt30)=0.84$

$θ = arccos (0.84) = {32.8}^{\circ}$ $theta=arccos(0.84)=32.8^@$

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What is the angle between $< - 1, 0, 6 >$ $<-1,0,6 >$ and $< 2, - 1, 5 >$ $< 2,-1,5 >$ ?

1 Answer

Explanation:

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