Question

Vector question. Let v=<-2,1> u=<3,-5> , find:

1)v * u
2)2)Find the angle formed between v and u

Answer 1

1. )
`\vec {v}.\vec{u}= -11`

2.) The angle between v and u is 147.52°.

Explanation:

Given:

`\vec {v}= ( -2 , 1 )\\\vec {u}= ( 3 , -5 )`

To Find:

1.
`\vec {v}.\vec{u}= ?`

2.
`\theta = ?`

Solution:

`\vec {v}.\vec{u}` is scalar product given as,

`\vec {v}.\vec{u}=|\vec {v}||\vec {u}|\cos \theta`

`|\vec {v}|=|(-2, 1)|=\sqrt{(-2)^(2) +1^(2)}=√(5)\\|\vec {u}|=|(3, 5)|=\sqrt{(3)^(2) +(-5)^(2)}=√(34)`

`\vec {v}.\vec{u}=(-2i +j).(3i-5j)\\`

Here only i.i = j.j =1 and i.j = j.i = 0

∴
`\vec {v}.\vec{u}=(-2* 3 +1* -5)\\\vec {v}.\vec{u}=-6-5=-11 \\`

Now, Substituting the above values we get

`-11=√(5)* √(34)\cos \theta\\ \cos \theta=(-11)/(√(170)) \\ \cos \theta =-0.84366\\\therefore \theta =cos^(-1)(-0.84366)\\\therefore \theta =147.52\°`

As it is negative mean
`\theta` is in Second Quadrant Because Cosine is negative in Second Quadrant.

1. )
`\vec {v}.\vec{u}= -11`

2.) The angle between v and u is 147.52°.