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Given the two vectors below, find the angle between them. Round answers to the nearest hundredth. v=⟨−13,−1⟩v=⟨−13,−1⟩ w=⟨19,−3.1⟩

Guest Mar 28, 2017
#1
+87293
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cos (theta)  =   [ u dot v ] /  [ length of u * length of v ]

u  = (-13, -1)   v   = (19, -3.1)

u dot v  =  ( -13 * 19  +   -1 * -3.1)  =  (-247 + 3.1)  = -243.9

Length of u   =   sqrt[ 13^2 + 1^1) =  sqrt (170)

Length of v  = sqrt ( 19^2 + 3.1^2)   = sqrt (370.61)

cos (theta  =   -243.9 / [ sqrt (170) * sqrt ( 370.61) ]

arccos  [ -243.9 / [ sqrt (170) * sqrt (370.61) ] = theta  ≈  166.33°

CPhill  Mar 28, 2017
#2
+87293
0

Sorry.....the vectors should be noted as v and w, not u and v....but......the the same procedure holds

CPhill  Mar 28, 2017
#4
+19599
+4

Given the two vectors below, find the angle between them. Round answers to the nearest hundredth.

v=⟨−13,−1⟩ w=⟨19,−3.1⟩

$$\vec{v} = \binom{-13}{-1} \\ \vec{w} = \binom{19}{-3.1}$$

$$\begin{array}{|rcll|} \hline \tan(\varphi) &=& \frac{ |\vec{v} \times \vec{w}| } {\vec{v} \cdot \vec{w}} \\ &=& \frac{ \left|\binom{-13}{-1} \times \binom{19}{-3.1} \right| } {\binom{-13}{-1} \cdot \binom{19}{-3.1} } \\ &=& \frac{ (-13)\cdot(-3.1)-(-1)\cdot 19 } {(-13)\cdot 19 + (-1)\cdot(-3.1) } \\ &=& \frac{ 40.3+19 } {-247 + 3.1 } \\ &=& \frac{ 59.3 } {-243.9 } \quad & | \quad \frac{+}{-}\ \text{Quadrant }\ II. \\ &=& -\frac{ 59.3 } {243.9 } \\ \tan(\varphi) &=& -0.24313243132 \\ \varphi &=& \arctan( -0.24313243132 ) + 180^{\circ} \\ &=& -13.6653125958+ 180^{\circ} \\ \varphi &=& 166.334687404^{\circ} \\ \hline \end{array}$$

The angle between $$\vec{v}$$ and $$\vec{w}$$ is $$166.33^{\circ}$$

heureka  Mar 28, 2017