Answer:
[tex]\theta = 108.29[/tex]
Step-by-step explanation:
Given
[tex]u = <2,6>[/tex]
[tex]v = <4,-7>[/tex]
Required:
Calculate the angle between u and v
The angle [tex]\theta[/tex] is calculated as thus:
[tex]cos\theta = \frac{u.v}{|u|.|v|}[/tex]
For a vector
[tex]A = <a,b>[/tex]
[tex]A = a * b[/tex]
[tex]cos\theta = \frac{u.v}{|u|.|v|}[/tex] becomes
[tex]cos\theta = \frac{<2,6>.<4,-7>}{|u|.|v|}[/tex]
[tex]cos\theta = \frac{2*6+4*-7}{|u|.|v|}[/tex]
[tex]cos\theta = \frac{12-28}{|u|.|v|}[/tex]
[tex]cos\theta = \frac{-16}{|u|.|v|}[/tex]
For a vector
[tex]A = <a,b>[/tex]
[tex]|A| = \sqrt{a^2 + b^2}[/tex]
So;
[tex]|u| = \sqrt{2^2 + 6^2}[/tex]
[tex]|u| = \sqrt{4 + 36}[/tex]
[tex]|u| = \sqrt{40}[/tex]
[tex]|v| = \sqrt{4^2+(-7)^2}[/tex]
[tex]|v| = \sqrt{16+49}[/tex]
[tex]|v| = \sqrt{65}[/tex]
So:
[tex]cos\theta = \frac{-16}{|u|.|v|}[/tex]
[tex]cos\theta = \frac{-16}{\sqrt{40}*\sqrt{65}}[/tex]
[tex]cos\theta = \frac{-16}{\sqrt{2600}}[/tex]
[tex]cos\theta = \frac{-16}{\sqrt{100*26}}[/tex]
[tex]cos\theta = \frac{-16}{10\sqrt{26}}[/tex]
[tex]cos\theta = \frac{-8}{5\sqrt{26}}[/tex]
Take arccos of both sides
[tex]\theta = cos^{-1}(\frac{-8}{5\sqrt{26}})[/tex]
[tex]\theta = cos^{-1}(\frac{-8}{5 * 5.0990})[/tex]
[tex]\theta = cos^{-1}(\frac{-8}{25.495})[/tex]
[tex]\theta = cos^{-1}(-0.31378701706)[/tex]
[tex]\theta = 108.288386087[/tex]
[tex]\theta = 108.29[/tex] (approximated)
