If $A = < 4, 1, 5 >$ $A = <4 ,1 ,5 >$ , $B = < 6, 2, - 2 >$ $B = <6 ,2 ,-2 >$ and $C = A - B$ $C=A-B$ , what is the angle between A and C?

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If $A = < 4, 1, 5 >$ $A = <4 ,1 ,5 >$ , $B = < 6, 2, - 2 >$ $B = <6 ,2 ,-2 >$ and $C = A - B$ $C=A-B$ , what is the angle between A and C?

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Answer 1 · 2017-11-19T13:34:09

The angle is $= {68.2}^{\circ}$ $=68.2^@$

Explanation:

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

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

Where $θ$ $theta$ is the angle between $\to A$ $vecA$ and $\to C$ $vecC$

The dot product is

$\to A . \to C = ⟨ 4, 1, 5 ⟩ . ⟨ 6, 2, - 2 ⟩ = 24 + 2 - 10 = 16$ $vecA.vecC=〈4,1,5〉.〈6,2,-2〉=24+2-10=16$

The modulus of $\to A$ $vecA$ = $∥ ∥ ⟨ 4, 1, 5 ⟩ ∥ = \sqrt{16 + 1 + 25} = \sqrt{42}$ $∥〈4,1,5〉∥=sqrt(16+1+25)=sqrt42$

The modulus of $\to C$ $vecC$ = $∥ ∥ ⟨ 6, 2, - 2 ⟩ ∥ = \sqrt{36 + 4 + 4} = \sqrt{44}$ $∥〈6,2,-2〉∥=sqrt(36+4+4)=sqrt44$

So,

$cos θ = \frac{\to A . \to C}{∥ ∥ ∥ \to A ∥ \cdot ∥ \to C ∥ ∥ ∥} = \frac{16}{\sqrt{42} \cdot \sqrt{44}} = 0.37$ $costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=16/(sqrt42*sqrt44)=0.37$

$θ = {68.2}^{\circ}$ $theta=68.2^@$

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If $A = < 4, 1, 5 >$ $A = <4 ,1 ,5 >$ , $B = < 6, 2, - 2 >$ $B = <6 ,2 ,-2 >$ and $C = A - B$ $C=A-B$ , what is the angle between A and C?

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Explanation:

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If A=<4,1,5>A = <4 ,1 ,5 >, B=<6,2,−2>B = <6 ,2 ,-2 > and C=A−BC=A-B, what is the angle between A and C?

1 Answer

Explanation:

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If $A = < 4, 1, 5 >$ $A = <4 ,1 ,5 >$ , $B = < 6, 2, - 2 >$ $B = <6 ,2 ,-2 >$ and $C = A - B$ $C=A-B$ , what is the angle between A and C?