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?

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

We're asked to find the angle between vector `vecb` 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 `veci`, would be, two dimensionally,

`veci = 1hati + 0hatj`

The dot product of these two vectors is given by

`vecb • veci = bicosphi`

where

• `b` is the magnitude of `vecb`

• `i` is the magnitude of `veci`

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

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

`phi = arccos((vecb • veci)/(bi))`

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

The dot product is

`vecb • veci = b_x i_x + b_yi_y = (-17.8)(1) + (5.1)(0) = color(red)(-17.8`

The magnitude of each vector is

`b = sqrt((b_x)^2 + (b_y)^2) = sqrt((-17.8)^2 + (5.1)^2) = 18.5`

`i = sqrt((i_x)^2 + (i_y)^2) = sqrt((1)^2 + (0)^2) = 1`

Thus, the angle between the vectors is

`phi = arccos((-17.8)/((18.5)(1))) = color(blue)(164^"o"`

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 `vecb` to find the angle measured anticlockwise from the positive `x`-axis:

`phi = arctan((5.1)/(-17.8)) = -16.0^"o"`

We must add `180^"o"` to this angle due to the calculator error; `vecb` is actually in the second quadrant:

`-16.0^"o" + 180^"o" = color(blue)(164^"o"`