Question

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

E <6, 4>, v <7, 5
O 0.9°
O1.8%
O-9.19
O 11.8

Answer 1

α=arctan(5/7)≈35.5°

ß=arctan(4/6)≈33.7°

α-ß≈1.8°

Answer 2

1.8°

Explanation:

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

Magnitude of u

`|u|=√(6^2+4^2)\\\Rightarrow |u|=√(36+16)\\\Rightarrow |u|=\sqrt {52}`

Magnitude of v

`|v|=√(7^2+5^2)\\\Rightarrow |u|=√(49+25)\\\Rightarrow |u|=\sqrt {74}`

`cos\alpha =(u.v)/(|u|\ |v|)\\\Rightarrow cos\alpha =\frac{6* 7+4* 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°