Answer:
The angle between the given vectors to the nearest tenth of a degree is:
6 degree
Step-by-step explanation:
Let α be the angle between two vectors.
The cosine of the angle between two vectors is given by:
\(\cos \alpha=\dfrac{u\cdot v}{|u||v|}\)
Here we have vector u as:
\(u=<2,-4>\)
and
\(v=<3,-8>\)
\(u\cdot v=<2,-4>\cdot <3,-8>\\\\u\cdot v=2\times 3+(-4)\times (-8)\\\\u\cdot v=6+32\\\\u\cdot v=38\)
Also,
\(|u|=\sqrt{2^2+(-4)^2}\\\\|u|=\sqrt{4+16}\\\\|u|=\sqrt{20}\\\\|u|=2\sqrt{5}\)
and
\(|v|=\sqrt{3^2+(-8)^2}\\\\|v|=\sqrt{9+64}\\\\|v|=\sqrt{73}\)
Hence, we get:
\(\cos \alpha=\dfrac{38}{2\sqrt{5}\cdot \sqrt{73}}\\\\\cos \alpha=0.9945\\\\i.e.\\\\\alpha=\arccos 0.9945\\\\\alpha=6.01198\)
which to the nearest tenth of a degree is: 6°
