What is the angle between $< 4, 1, - 2 >$ $< 4 , 1 , -2 >$ and $< - 9, - 3, 2 >$ $< -9 , -3 , 2 >$ ?

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What is the angle between $< 4, 1, - 2 >$ $< 4 , 1 , -2 >$ and $< - 9, - 3, 2 >$ $< -9 , -3 , 2 >$ ?

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Answer 1 · 2018-03-12T17:52:28

The angle is $= {165.4}^{\circ}$ $=165.4^@$

Explanation:

The angle between $2$ $2$ vectors, $\to A$ $vecA$ 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

$\to A . \to B = ⟨ 4, 1, - 2 ⟩ . ⟨ - 9, - 3, 2 ⟩ = (4) \cdot (- 9) + (1) \cdot (- 3) + (- 2) \cdot (2) = - 43$ $vecA.vecB=〈4,1,-2〉.〈-9,-3,2〉=(4)*(-9)+(1)*(-3)+(-2)*(2)=-43$

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

The modulus of $\to B$ $vecB$ = $∥ ∥ ⟨ - 9, - 3, 2 ⟩ ∥ = \sqrt{81 + 9 + 4} = \sqrt{94}$ $∥〈-9,-3,2〉∥=sqrt(81+9+4)=sqrt94$

So,

$cos θ = \frac{\to A . \to B}{∥ ∥ ∥ \to A ∥ \cdot ∥ \to B ∥ ∥ ∥} = - \frac{43}{\sqrt{21} \cdot \sqrt{94}} = - 0.97$ $costheta=(vecA.vecB)/(∥vecA∥*∥vecB∥)=-43/(sqrt21*sqrt94)=-0.97$

$θ = arccos (- 0.97) = {165.4}^{\circ}$ $theta=arccos(-0.97)=165.4^@$

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What is the angle between $< 4, 1, - 2 >$ $< 4 , 1 , -2 >$ and $< - 9, - 3, 2 >$ $< -9 , -3 , 2 >$ ?

1 Answer

Explanation:

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What is the angle between < 4 , 1 , -2 > <4,1,−2>< 4 , 1 , -2 > and < -9 , -3 , 2 > <−9,−3,2> < -9 , -3 , 2 > ?

1 Answer

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What is the angle between $< 4, 1, - 2 >$ $< 4 , 1 , -2 >$ and $< - 9, - 3, 2 >$ $< -9 , -3 , 2 >$ ?