If $A = < 1, 6, 9 >$ $A = <1 ,6 ,9 >$ , $B = < - 9, 4, - 8 >$ $B = <-9 ,4 ,-8 >$ and $C = A - B$ $C=A-B$ , what is the angle between A and C?

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Answer 1 · 2016-11-15T16:38:30

The angle is $= 35.6$ $=35.6$ º

Let's start by calculating $\to C$ $vecC$

$\to C = \to A - \to B = ⟨ 1, 6, 9 ⟩ - ⟨ - 9, 4, - 8 ⟩ = ⟨ 10, 2, 17 ⟩$ $vecC=vecA-vecB=〈1,6,9〉-〈-9,4,-8〉=〈10,2,17〉$

The angle $θ$ $theta$ is given by the dot product definition

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

$\to A . \to C = ⟨ 10, 2, 17 ⟩ . ⟨ 1, 6, 9 ⟩ = 10 + 12 + 153 = 175$ $vecA.vecC=〈10,2,17〉.〈1,6,9〉=10+12+153=175$

The modulus of $\to A = ∥ \to A ∥ = ∥ ⟨ 1, 6, 9 ⟩ ∥ = \sqrt{1 + 36 + 81} = \sqrt{118}$ $vecA=∥vecA∥=∥〈1,6,9〉∥=sqrt(1+36+81)=sqrt118$

The modulus of $\to C = ∥ \to C ∥ = ∥ ⟨ 10, 2, 17 ⟩ ∥ = \sqrt{100 + 4 + 289} = \sqrt{393}$ $vecC=∥vecC∥=∥〈10,2,17〉∥=sqrt(100+4+289)=sqrt393$

Therefore, $cos θ = \frac{175}{\sqrt{118} \cdot \sqrt{393}} = 0.81$ $costheta=175/(sqrt118*sqrt393)=0.81$

$θ = 35.6$ $theta=35.6$ º

If A=<1,6,9>A = <1 ,6 ,9 >, B=<−9,4,−8>B = <-9 ,4 ,-8 > and C=A−BC=A-B, what is the angle between A and C?