Question

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

Answer 1

Answer:

1.8°

Step-by-step explanation:

u = <6,4> , v = <7,5>

Magnitude of u

$|u|=\sqrt{6^2+4^2}\\\Rightarrow |u|=\sqrt{36+16}\\\Rightarrow |u|=\sqrt {52}$

Magnitude of v

$|v|=\sqrt{7^2+5^2}\\\Rightarrow |u|=\sqrt{49+25}\\\Rightarrow |u|=\sqrt {74}$

$cos\alpha =\frac{u.v}{|u|\ |v|}\\\Rightarrow cos\alpha =\frac{6\times 7+4\times 5}{\sqrt {52} \sqrt {74}}\\\Rightarrow cos\alpha =0.99948\\\Rightarrow \alpha =cos^{-1}0.99948\\\Rightarrow \alpha =1.847^{\circ}$

∴ Angle between the vectors is 1.8°

Answer 2

Α=arctan(5/7)≈35.5°

ß=arctan(4/6)≈33.7°

α-ß≈1.8°