+280 this is the second answer i got
16.865987621884, is it correct. i cant afford to keep puttin in wrong answers. If its the answer then i know what im doing now.
+128406 u = < - 2, 8 >
v = < -7, - 9 >
cos (theta) = u (dot) v / [ ll u ll * ll v ll ]
u (dot) v = 14 - 72 = -58
ll u ll = sqrt [ (-2)^2 + (8)^2 ] = sqrt (68)
ll v ll = sqrt [ (-7)^2 + (-9)^2 ] = sqrt [ 49 + 81] = sqrt (130)
cos (theta) = -58 / [ sqrt (68) * sqrt (130) ]
arcos [ -58 / [ sqrt (68) * sqrt (130) ] ] = theta ≈ 128.0888°
Here is a graph that shows this, Veteran :
+26367 Find the angle between the vectors. State your answer in degrees,
rounded to at least four decimal places.
\(\vec{u} = \binom{-2}{8} \\ \vec{v} = \binom{-7}{-9} \\\)
\(\begin{array}{|rcll|} \hline \tan(\theta) &=& \frac{|~\vec{u} \times \vec{v}~| } {\vec{u} \cdot \vec{v} } \\ &=& \frac{ \left|~\binom{-2}{8} \times \binom{-7}{-9}~\right| } {\binom{-2}{8} \cdot \binom{-7}{-9} } \\ &=& \frac{ (-2)\cdot (-9) - (8)\cdot (-7) } { (-2)\cdot (-7) + (8)\cdot (-9) } \\ &=& \frac{ 18+56 } { 14-72 } \\ &=& \frac{ 74 } { -58 } \quad & | \quad II.\text{Quadrant} \\ &=& \frac{ 37 } { -29 } \\ \theta &=& \arctan(\frac{ 37 } { -29 }) \\ \theta &=& \arctan(-1.27586206897) \\ \theta &=& -51.9112271190180^{\circ} + 180^{\circ} \quad & | \quad II.\text{Quadrant} \\ \theta &=& 128.088772881^{\circ} \\ \theta &\approx& 128.0888^{\circ} \\ \hline \end{array}\)
