Question

Find the angle between the given vectors. Round your answer, in degrees, to two decimal places. u=⟨2,−6⟩u=⟨2,−6⟩, v=⟨4,−7⟩

Answer 1

`\theta = 108.29`

Explanation:

Given

`u = <2,6>`

`v = <4,-7>`

Required:

Calculate the angle between u and v

The angle
`\theta` is calculated as thus:

`cos\theta = (u.v)/(|u|.|v|)`

For a vector

`A = <a,b>`

`A = a * b`

`cos\theta = (u.v)/(|u|.|v|)` becomes

`cos\theta = (<2,6>.<4,-7>)/(|u|.|v|)`

`cos\theta = (2*6+4*-7)/(|u|.|v|)`

`cos\theta = (12-28)/(|u|.|v|)`

`cos\theta = (-16)/(|u|.|v|)`

For a vector

`A = <a,b>`

`|A| = √(a^2 + b^2)`

So;

`|u| = √(2^2 + 6^2)`

`|u| = √(4 + 36)`

`|u| = √(40)`

`|v| = √(4^2+(-7)^2)`

`|v| = √(16+49)`

`|v| = √(65)`

So:

`cos\theta = (-16)/(|u|.|v|)`

`cos\theta = (-16)/(√(40)*√(65))`

`cos\theta = (-16)/(√(2600))`

`cos\theta = (-16)/(√(100*26))`

`cos\theta = (-16)/(10√(26))`

`cos\theta = (-8)/(5√(26))`

Take arccos of both sides

`\theta = cos^(-1)((-8)/(5√(26)))`

`\theta = cos^(-1)((-8)/(5 * 5.0990))`

`\theta = cos^(-1)((-8)/(25.495))`

`\theta = cos^(-1)(-0.31378701706)`

`\theta = 108.288386087`

`\theta = 108.29` (approximated)