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⟩
+129847 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°
+129847 Sorry.....the vectors should be noted as v and w, not u and v....but......the the same procedure holds
+26387 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 \(166.33^{\circ}\)
