find the angle between the given vectors to the nearest tenth of degree, u =(4,2) and v =(5,4) 

2 Answers

llltkl's profile pic

llltkl | College Teacher | (Level 3) Valedictorian

Posted on

The angle between two nonzero vectors is the angle `theta` ,` 0lt=thetalt=pi` , between their

respective standard position vectors. This angle can be found using the dot product:

`u . v = ||u|| ||v|| cos theta`

`rArr cos theta = (u . v)/(||u|| ||v||) ` -----------(i)

Here, `u =(4,2)` and `v =(5,4)`  .

So, `u . v = (4)(5) + (2)(4) = 28`



Substituting the values in equation (i) we get:

`cos theta=(28)/(2sqrt5*sqrt(41))`

`rArr cos theta=(28)/(2sqrt(205))`

This implies that the angle between the two vectors is:

`theta=arccos ((28)/(2sqrt(205)))~~12.1^o`

kailash's profile pic

kailash | Elementary School Teacher | eNotes Newbie

Posted on

The position vector  `vec{ OU}=4veci+2vec j` .

The position vector `vec{OV}=5veci+4 vecj`  

Let `theta ` be the angle between the vectors OU and OV.

`vec{OU}.vec{OV}=|vec{OU}||vec{OV}| cos(theta)`


`|vec {OU}|=sqrt(4^2+2^2)=2sqrt(5)`





`28=2sqrt(5) sqrt(41)cos(theta)`