What is the angle between $< 2, 9, - 2 >$ $<2,9,-2>$ and $< 0, 3, 0 >$ $<0,3,0 >$ ?

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What is the angle between $< 2, 9, - 2 >$ $<2,9,-2>$ and $< 0, 3, 0 >$ $<0,3,0 >$ ?

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Answer 1 · 2017-03-03T11:10:05

The angle is $= 17.4$ $=17.4$ 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

$\to A . \to B = ⟨ 2, 9, - 2 ⟩ . ⟨ 0, 3, 0 ⟩ = 27$ $vecA.vecB=〈2,9,-2〉.〈0,3,0 〉 = 27$

The modulus of $\to A$ $vecA$ = $∥ ∥ ⟨ 2, 9, - 2 ⟩ ∥ = \sqrt{4 + 81 + 4} = \sqrt{89}$ $∥〈2,9,-2〉∥=sqrt(4+81+4)=sqrt89$

The modulus of $\to B$ $vecB$ = $∥ ∥ ⟨ 0, 3, 0 ⟩ ∥ = \sqrt{9} = 3$ $∥〈0,3,0〉∥=sqrt9=3$

So,

$cos θ = \frac{\to A . \to B}{∥ ∥ ∥ \to A ∥ \cdot ∥ \to B ∥ ∥ ∥} = \frac{27}{\sqrt{89} \cdot 3} = 0.95$ $costheta=(vecA.vecB)/(∥vecA∥*∥vecB∥)=27/(sqrt89*3)=0.95$

$θ = 17.4$ $theta=17.4$ º

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What is the angle between $< 2, 9, - 2 >$ $<2,9,-2>$ and $< 0, 3, 0 >$ $<0,3,0 >$ ?

1 Answer

Explanation:

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What is the angle between <2,9,-2> <2,9,−2><2,9,-2> and <0,3,0 ><0,3,0><0,3,0 >?

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What is the angle between $< 2, 9, - 2 >$ $<2,9,-2>$ and $< 0, 3, 0 >$ $<0,3,0 >$ ?