What is the angle between $< 4, 7, 1 >$ $<4,7,1 >$ and $< 9, 5, 4 >$ $<9,5,4>$ ?

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Answer 1 · 2016-11-12T19:21:25

The angle is $= 33.3$ $=33.3$ º

Let the angle be $θ$ $theta$

Let $\to u = ⟨ 4, 7, 1 ⟩$ $vecu=〈4,7,1〉$

and $\to v = ⟨ 9, 5, 4 ⟩$ $vecv=〈9,5,4〉$

The angle is given by the dot product definition

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

$\to u . \to v = ⟨ 4, 7, 1 ⟩ . ⟨ 9, 5, 4 ⟩ = 36 + 35 + 4 = 75$ $vecu.vecv=〈4,7,1〉.〈9,5,4〉=36+35+4=75$

$∥ ∥ \to u ∥ = ∥ ⟨ 4, 7, 1 ⟩ ∥ = \sqrt{16 + 49 + 1} = \sqrt{66}$ $∥vecu∥=∥〈4,7,1〉∥=sqrt(16+49+1)=sqrt66$

$∥ ∥ \to v ∥ = ∥ ⟨ 9, 5, 4 ⟩ ∥ = \sqrt{81 + 25 + 16} = \sqrt{122}$ $∥vecv∥=∥〈9,5,4〉∥=sqrt(81+25+16)=sqrt122$

Therefore, $cos θ = \frac{75}{\sqrt{66} \cdot \sqrt{122}} = 0.84$ $costheta=75/(sqrt66*sqrt122)=0.84$

$θ = 33.3$ $theta=33.3$ º

What is the angle between <4,7,1><4,7,1 > and <9,5,4><9,5,4> ?