Question #89959

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Question #89959

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Answer 1 · 2017-03-22T09:28:57

The angle is $= 41$ $=41$ º

Explanation:

The angle between $\to A$ $vecA$ and $\to B$ $vecB$ is given by the dot product definition.

$\to A . \to B = ∥ \to A ∥ \cdot ∥ \to B ∥ cos θ$ $vecA.vecB=∥vecA∥*∥vecB∥costheta$

Where $θ$ $theta$ is the angle between $\to A$ $vecA$ and $\to B$ $vecB$

The dot product is

$\to A . \to B = ⟨ 2, 3, 1 ⟩ . ⟨ 4, 1, 2 ⟩ = 8 + 3 + 2 = 13$ $vecA.vecB=〈2,3,1〉.〈4,1,2〉=8+3+2=13$

The modulus of $\to A$ $vecA$ = $∥ ∥ ⟨ 2, 3, 1 ⟩ ∥ = \sqrt{4 + 9 + 1} = \sqrt{14}$ $∥〈2,3,1〉∥=sqrt(4+9+1)=sqrt14$

The modulus of $\to B$ $vecB$ = $∥ ∥ ⟨ 4, 1, 2 ⟩ ∥ = \sqrt{16 + 1 + 4} = \sqrt{21}$ $∥〈4,1,2〉∥=sqrt(16+1+4)=sqrt21$

So,

$cos θ = \frac{\to A . \to B}{∥ ∥ ∥ \to A ∥ \cdot ∥ \to B ∥ ∥ ∥} = \frac{13}{\sqrt{14} \cdot \sqrt{21}} = 0.758$ $costheta=(vecA.vecB)/(∥vecA∥*∥vecB∥)=13/(sqrt14*sqrt21)=0.758$

$θ = 41$ $theta=41$ º

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Question #89959

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