Question

Find the angle between the given vectors to the nearest tenth of a degree.

u = <-5, -4>, v = <-4, -3>

-9.1°

1.8°

0.9°

11.8°

Answer 1

Dot product of vectors is given by:
\[\bar u· \bar v=u_x·v_x+u_y·v_y=\left| \bar u \right|\left| \bar v \right|\cos(\theta)\]With this, it is easy to calculate \[\cos(\theta)=\frac{ u_x·v_x+u_y·v_y }{ \left| \bar u \right| \left| \bar v \right|}\]

Answer 2

\[\cos(\theta)=\frac{ u_xv_x+u_yv_y }{ \sqrt{u_x^2+u_y^2}\sqrt{v_x^2+v_y^2} }\]In your case\[\cos(\theta)=\frac{ (-5)(-4)+(-4)(-3) }{ \sqrt{25+16}\sqrt{16+9}}=0.9995\rightarrow \theta=acos(0.9995)=1.8º\]