What is the angle between $< - 9, 2, 2 >$ $<-9 , 2 , 2 >$ and $< 5, 8, 3 >$ $< 5 , 8 , 3 >$ ?

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What is the angle between $< - 9, 2, 2 >$ $<-9 , 2 , 2 >$ and $< 5, 8, 3 >$ $< 5 , 8 , 3 >$ ?

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Answer 1 · 2018-04-29T15:41:29

The angle is $= {103.6}^{\circ}$ $=103.6^@$ anticlockwise from the x-axis

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

$\to A . \to B = ⟨ - 9, 2, 2 ⟩ . ⟨ 5, 8, 3 ⟩ = - 45 + 16 + 6 = - 22$ $vecA.vecB=〈-9,2,2〉.〈5,8,3〉=-45+16+6=-22$

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

The modulus of $\to C$ $vecC$ = $∥ ∥ ⟨ 5, 8, 3 ⟩ ∥ = \sqrt{25 + 64 + 9} = \sqrt{98}$ $∥〈5,8,3〉∥=sqrt(25+64+9)=sqrt98$

So,

$cos θ = \frac{\to A . \to B}{∥ ∥ ∥ \to A ∥ \cdot ∥ \to B ∥ ∥ ∥} = - \frac{22}{\sqrt{89} \cdot \sqrt{98}} = - 0.24$ $costheta=(vecA.vecB)/(∥vecA∥*∥vecB∥)=-22/(sqrt89*sqrt98)=-0.24$

$θ = arccos (- 0.24) = {103.6}^{\circ}$ $theta=arccos(-0.24)=103.6^@$ anticlockwise from the x-axis

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What is the angle between $< - 9, 2, 2 >$ $<-9 , 2 , 2 >$ and $< 5, 8, 3 >$ $< 5 , 8 , 3 >$ ?

1 Answer

Explanation:

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