How Is the Angle Calculated Between Displacement Vectors in Navigation?

[physics19]

A person walks in the following pattern: 6.6 km north, then 3.0 km west, and finally 7.0 km south. How far and in what direction would a bird fly in a straight line from the same starting point to the same final point?

I found that the distance would be 3.02 km. My problem is trying to find the angle. I came up with an angle of 7.59 degrees but that was not right. Please, any help would be appreciated.

 

[KEØM]

Try using this formula: cos($$\alpha$$) = (u(dot)v)$$/$$(|u||v|)

Sorry I am not very good with latex but it says that the cosine of alpha is equal to the dot product of the two vectors that you need the angle between over the magnitude of of one of your vectors times the other. So with your info solve for alpha.
If you need clarification please ask.

 

[baba89]

The angle between two vectors can be calculated using the dot product formula: θ = cos^-1 ((a · b) / (|a||b|)), where a and b are the two vectors. In this case, the vectors would be the displacement of the person (6.6 km north, 3.0 km west, and 7.0 km south) and the displacement of the bird (3.02 km straight line).

To find the angle, we first need to find the magnitude of the person's displacement vector, which can be calculated using the Pythagorean theorem. So, the magnitude would be √((6.6)^2 + (3.0)^2 + (7.0)^2) = 10.48 km.

Next, we need to find the magnitude of the bird's displacement vector, which is already given as 3.02 km.

Now, we can plug in these values into the dot product formula: θ = cos^-1 ((6.6*0 + 3.0*(-3.02) + 7.0*0) / (10.48*3.02)) = cos^-1 (-0.086) = 94.86 degrees.

Therefore, the angle between the person's displacement and the bird's displacement would be 94.86 degrees, which is the direction the bird would have flown in a straight line from the starting point to the final point.