Position and Displacement Vectors- The Watermelon Seed Vector

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Pmand92
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A watermelon seed has the following coordinates: x = -6.6 m, y = 2.2 m, and z = 0 m. Find its position vector as (a) a magnitude and (b) an angle relative to the positive direction of the x axis. If the seed is moved to the xyz coordinates (8.2 m, 0 m, 0 m), what is its displacement as (c) a magnitude and (d) an angle relative to the positive direction of the x axis?


A) Magnitude was found using (-6.6)^2+(2.2)^2=sqrt(48.4)=6.96m or 7.0m

B) The angle was found using inverse tangent of (2.2/6.6)=18.4 degrees. I subtracted 18.4 from 180 to get 161.6.

C) The magnitude was found using the displacement equation (Δx=X2-X1). 8.2m-(-6.6m)=14.8m

Up until part D, all of the answers are correct in WileyPlus. I have tried multiple ways trying to find this angle however, no success. Perhaps I am looking past the basic point. Please help.
 
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Hi, Pmand92. Can you show us the details of what you did for part (d)? What did you get for an answer?
 
Well I tried using the magnitude (14.8 squared) minus 2.2 squared to find a component so I could try the inverse tangent to find the angle. tan^-1(2.2/14.8)=8.5 degrees and then I subtract that from 180, which 171.5. Then at this point, I tried to put it into WileyPlus, which was wrong. So I tried inverse sin with just 2.2/14.8, which didn't work. At this point, I have switched out numbers just trying to see if I could get an answer for the WileyPlus to take the answer.
 
To be honest, I can't remember all of the ways to find the angle.
 
The displacement is a vector pointing from the initial position to the final position. Draw a sketch to see how that vector is oriented relative to the positive x direction.
 
Draw an x-y cartesian coordinate system and locate the initial and final locations (positions) of the seed. The displacement will be a vector pointing from the initial location to the final location. You don't need to worry about the z-coordinate since it's 0 for both the intial and final postitions.