What is the Displacement and Force Acting on a Particle in the xy Plane?

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Homework Statement



A particle moving in the xy plane undergoes a displacement ⃗s = (sx ˆi + sy ˆj), with sx = 1.74 m, sy = 4.3 m, while a constant force
F⃗ = (Fxˆi+Fyˆj), with Fx = 5.1 N, Fy = 1.95 N, acts on the particle.
Calculate the magnitude of the displacement.
Answer in units of m.

Find the magnitude of the force. Answer in units of N.

Calculate the work done by F⃗ . Answer in units of J.

Calculate the angle between F⃗ and ⃗s. Answer in units of degrees.

Homework Equations



The Attempt at a Solution



i took the dot product of the two thinking i would get the magnitude of the displacement. I am lost as to why it is incorrect?

1.74*5.1 + 4.3*1.95 = 17.259
 
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its simple, i figured it out... i feel like an idiot

pythagorean theorem, magnitude = sqrt(1.74^2 + 4.3^2)
 
the entire problem is simple, i figured it out so i thought i would post it in case anyone would stumble upon this in the future

Find the magnitude of the force. Answer in units of N.

this is just the length force vector, duh...

Calculate the work done by F⃗ . Answer in units of J.

this is the dot product of the two

Calculate the angle between F⃗ and ⃗s. Answer in units of degrees.

this angle is found by taking the inverse of any trig function, provided that you use the correct sides that coorespond to that function, of the larger angle, - the smaller angle.