Find the angle between v=4i+2j and w=2i-6j

Question

I have one question on my precalculus homework that i can't understamd. Can you please solve it step by step? 

Answer 1

This is a vector problem. 

 

You know for the first vector if you draw it on the coordinate plane axis.  It goes +4 in the x direction and +2 in the y direction. 

 

For the 2nd vector it goes 2 in the x direction and -6 in the y direction. 

 

We already have x and y components for each, so we can put it in this notation v=<4,2> and w=<2,-6>

 

 

 

Basically what we have to do is find the dot product of each first. 

 

So the dot product is 4*2 + 2*-6 = -4

 

Then we have to find the magnitude of each of the vectors which is the square root of the the  first vector squared and added together then square rooted.  Then the square root of the y components added together and squared. 

 

So we do this :

 

√4²+2² which comes out to √20.  You have to do this to vector w as well which is √2²+(-6)² = √40

 

The formula to find the angle between the vectors is

 

 

cos Θ= -4/(√20 x √40).  You then take an inverse cosine of this to find the angle.  It comes out to approximately 98 degrees.