How do you find the angel between u=<-6,-2> and v=<2,12>?

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Answer 1 · 2016-11-21T12:49:43

The angle is $117.9$ $117.9$ º

The angle is given by the dot product definition

$\to u . \to v = ∥ \to u ∥ \cdot ∥ \to v ∥ \cdot cos θ$ $vecu.vecv=∥vecu∥*∥vecv∥*costheta$

Where $θ$ $theta$ is the angle between the 2 vectors.

The dot prduct is $\to u . \to v = ⟨ - 6, - 2 ⟩ . ⟨ 2, 12 ⟩ = (- 12 - 24) = - 36$ $vecu.vecv =〈-6,-2〉.〈2,12〉=(-12-24)=-36$

The modulus of $\to u = ∥ ⟨ - 6, - 2 ⟩ ∥ = \sqrt{34 + 4} = \sqrt{40}$ $vecu=∥〈-6,-2〉∥=sqrt(34+4)=sqrt40$

The modulus of $\to v = ∥ ⟨ 2, 12 ⟩ ∥ = \sqrt{4 + 144} = \sqrt{148}$ $vecv=∥〈2,12〉∥=sqrt(4+144)=sqrt148$

Therefore, #costheta=vecu.vecv/(∥vecu∥*∥vecv∥)#

$= - \frac{36}{\sqrt{40} \cdot \sqrt{148}} = - 0.47$ $=-36/(sqrt40*sqrt148)=-0.47$

$θ = 117.9$ $theta=117.9$ º