Displacement of a diving dolphin

[brundlfly]

Homework Statement

a dolphin swims near the sea surface 350m west, then 200m northwest, and then dives below the surface and swims an additional distance in an unknown direction. The final position of the dolphin is 80m directly below the fishing boat located 400m north of the starting point. Calculate the dolphin's displacement during the dive.

Homework Equations

-Conversion of polar coordinates to Cartesian coordinates

The Attempt at a Solution

I converted the start point of the dive to A = 141(-i) + 141(j) and the position of the boat as B = 0(i) + 400(j). I added A and B together so I got -141(i) + 541(j) + 0(k), then I added the vector of the position of the dolphin 80 underneath the boat which is 0(i)+0(j)-80(k) and got
-141(i)+541(j)-80(k). Then I calculated the norm of the vector to get the displacement of the dolphin during the dive: sqrt(141^2+541^2+80^2) and as a result I got 564.7672 meters. None of the calculations involve the other vectors, so my question is if I have to use them since only the displacement of the dive is needed. Am I missing something? Thank you for your time and effort.

 

[NihalSh]

I converted the start point of the dive to A = 141(-i) + 141(j)

this is not the diving point. diving point is the sum of two vectors - 350m west, then 200m northwest
convert it in vector form and you should arrive at the correct answer.
the procedure is correct.