Two vectors are given by a = 3.3 x - 6.4 y and b = -17.8 x + 5.1 y. What is the angle between vector b and the positive x-axis?

1 Answer
Jul 8, 2017

ϕ=164o

Explanation:

Here's a more rigorous way to do this (easier way at the bottom):

We're asked to find the angle between vector b and the positive x-axis.

We'll imagine there is a vector that points in the positive x-axis direction, with magnitude 1 for simplifications. This unit vector, which we'll call vector i, would be, two dimensionally,

i=1ˆi+0ˆj

The dot product of these two vectors is given by

bi=bicosϕ

where

  • b is the magnitude of b

  • i is the magnitude of i

  • ϕ is the angle between the vectors, which is what we're trying to find.

We can rearrange this equation to solve for the angle, ϕ:

ϕ=arccosbibi

We therefore need to find the dot product and the magnitudes of both vectors.

The dot product is

bi=bxix+byiy=(17.8)(1)+(5.1)(0)=17.8

The magnitude of each vector is

b=(bx)2+(by)2=(17.8)2+(5.1)2=18.5

i=(ix)2+(iy)2=(1)2+(0)2=1

Thus, the angle between the vectors is

ϕ=arccos(17.8(18.5)(1))=164o

Here's an easier way to do this:

This method can be used since we're asked to find the angle between a vector and the positive x-axis, which is where we typically measure angles from anyway.

Therefore, we can simply take the inverse tangent of vector b to find the angle measured anticlockwise from the positive x-axis:

ϕ=arctan(5.117.8)=16.0o

We must add 180o to this angle due to the calculator error; b is actually in the second quadrant:

16.0o+180o=164o