If $A = < 6, 6, - 3 >$ $A = <6 ,6 ,-3 >$ , $B = < - 9, 6, - 2 >$ $B = <-9 ,6 ,-2 >$ and $C = A - B$ $C=A-B$ , what is the angle between A and C?

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Answer 1 · 2016-11-26T11:08:58

The angle is $= 46.6$ $=46.6$ º

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

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

$\to C = ⟨ 6, 6, - 3 ⟩ - ⟨ - 9, 6, - 2 ⟩ = ⟨ 15, 0, - 1 ⟩$ $vec C=〈6,6,-3〉-〈-9,6,-2〉=〈15,0,-1〉$

To calculate the angle $θ$ $theta$ between 2 vectors, we use the dot product definition

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

The dot product is $= ⟨ 6, 6, - 3 ⟩ . ⟨ 15, 0, - 1 ⟩ = 90 + 0 + 3 = 93$ $=〈6,6,-3〉.〈15,0,-1〉=90+0+3=93$

The modulus of $\to A = ∥ ⟨ 6, 6, - 3 ⟩ ∥ = \sqrt{36 + 36 + 9} = \sqrt{81} = 9$ $vecA=∥〈6,6,-3〉∥=sqrt(36+36+9)=sqrt81=9$

The modulus of $\to C = ∥ ⟨ 15, 0, - 1 ⟩ ∥ = \sqrt{225 + 1} = \sqrt{226}$ $vecC=∥〈15,0,-1〉∥=sqrt(225+1)=sqrt226$

So, $cos θ = \frac{93}{9 \sqrt{226}} = 0.69$ $costheta=93/(9sqrt226)=0.69$

And $θ = 46.6$ $theta=46.6$ º

If A = <6 ,6 ,-3 >A=<6,6,−3>A = <6 ,6 ,-3 >, B = <-9 ,6 ,-2 >B=<−9,6,−2>B = <-9 ,6 ,-2 > and C=A-BC=A−BC=A-B, what is the angle between A and C?