If $A = < - 5, 2, 1 >$ $A = <-5 ,2 ,1 >$ , $B = < - 8, 4, 9 >$ $B = <-8 ,4 ,9 >$ and $C = A - B$ $C=A-B$ , what is the angle between A and C?

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Answer 1 · 2016-11-17T10:20:52

The angle is $= 124.2$ $=124.2$ º

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

$\to C = \to A - \to B$ $vecC=vecA-vecB$

Therefore, $\to C = ⟨ - 5, 2, 1 ⟩ - ⟨ - 8, 4, 9 ⟩ = ⟨ 3, - 2, - 8 ⟩$ $vecC=〈-5,2,1〉-〈-8,4,9〉=〈3,-2,-8〉$

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

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

The dot product is
$\to A . \to C = ⟨ - 5, 2, 1 ⟩ . ⟨ 3, - 2, - 8 ⟩ = - 15 - 4 - 8 = - 27$ $vecA.vecC=〈-5,2,1〉.〈3,-2,-8〉=-15-4-8=-27$

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

The modulus of $\to C = ∥ ⟨ 3, - 2, - 8 ⟩ ∥ = \sqrt{9 + 4 + 64} = \sqrt{77}$ $vecC=∥〈3,-2,-8〉∥=sqrt(9+4+64)=sqrt77$

Therefore, $cos θ = \to A . \frac{\to C}{∥ ∥ ∥ \to A ∥ . ∥ \to C ∥ ∥ ∥}$ $costheta=vecA.vecC/(∥vecA∥.∥vecC∥)$

$= - \frac{27}{\sqrt{30} \cdot \sqrt{77}} = - 0.56$ $=-27/(sqrt30*sqrt77)=-0.56$

$θ = 124.2$ $theta=124.2$ º

If A=<−5,2,1>A = <-5 ,2 ,1 >, B=<−8,4,9>B = <-8 ,4 ,9 > and C=A−BC=A-B, what is the angle between A and C?