Question

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

Answer 1

Answer:

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

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

Step-by-step 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,

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

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

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

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

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

Now, Substituting the above values we get

$-11=\sqrt{5}\times \sqrt{34}\cos \theta\\ \cos \theta=\frac{-11}{\sqrt{170}} \\ \cos \theta =-0.84366\\\therefore \theta =cos^{-1}(-0.84366)\\\therefore \theta =147.52\°$

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°.