Question

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

Answer 1

Answer:

20.3

Step-by-step explanation:

Answer 2

Answer:

Step-by-step explanation:

u=6i-j

v=7i-4j

u.v=|6i-j||7i-4j| cos α,where α is the angle between u & v.

(6i-j).(7i-4j)=√(6²+(-1)²)√(7²+(-4)²) cos α

(6)(7)+(-1)(-4)=√37√65 cos α

42+4=√37√65  cos α

$\cos \alpha =\frac{46}{\sqrt{37} \sqrt{65} } \\\alpha =\cos^{-1}\frac{46}{\sqrt{37} \sqrt{65} } \approx 20.28^\circ \approx 20.3^\circ$