If $A = < 2, - 5, - 8 >$ $A = <2 ,-5 ,-8 >$ , $B = < - 9, 4, - 5 >$ $B = <-9 ,4 ,-5 >$ and $C = A - B$ $C=A-B$ , what is the angle between A and C?

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Answer 1 · 2016-11-15T08:44:57

The angle is $49.5$ $49.5$ º

Let's calculate $\to C$ $vecC$

$\to C = \to A - \to B = ⟨ 2, - 5, - 8 ⟩ - ⟨ - 9, 4, - 5 ⟩$ $vecC=vecA-vecB=〈2,-5,-8〉-〈-9,4,-5〉$

$= ⟨ 11, - 9, - 3 ⟩$ $=〈11,-9,-3〉$

The angle $θ$ $theta$ between $\to A$ $vecA$ and $\to C$ $vecC$ is calculated from the dot product definition

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

The dot product $\to A . \to C = ⟨ 2, - 5, - 8 ⟩ . ⟨ 11, - 9, - 3 ⟩$ $vecA.vecC=〈2,-5,-8〉.〈11,-9,-3〉$

$= 2 \cdot 11 + 5 \cdot 9 + 8 \cdot 3 = 22 + 45 + 24 = 91$ $=2*11+5*9+8*3=22+45+24=91$

The modulus of $∥ ∥ ∥ \to A ∥ = ∥ ⟨ 2, - 5, - 8 ⟩ ∥ = \sqrt{4 + 25 + 64} = \sqrt{93}$ $∥vecA∥=∥〈2,-5,-8〉∥=sqrt(4+25+64)=sqrt93$

The modulus of $∥ ∥ ∥ \to C ∥ = ∥ ⟨ 11, - 9, - 3 ⟩ ∥ = \sqrt{121 + 81 + 9} = \sqrt{211}$ $∥vecC∥=∥〈11,-9,-3〉∥=sqrt(121+81+9)=sqrt211$

Therefore, $cos θ = \frac{91}{\sqrt{93} \cdot \sqrt{211}} = 0.65$ $costheta=91/(sqrt93*sqrt211)=0.65$

$θ = 49.5$ $theta=49.5$ º

If A=<2,−5,−8>A = <2 ,-5 ,-8 >, B=<−9,4,−5>B = <-9 ,4 ,-5 > and C=A−BC=A-B, what is the angle between A and C?