Vectors A and B lie in the xy ‑plane. Vector A has a magnitude of 17.1 and is at an angle of 150.5∘ counterclockwise from the +x ‑axis. Vector B has a magnitude of 28.1 and is 205.3∘ from the +x ‑axis. Resolve A and B into components, and express using unit vectors, A =Ax+Ay+Az
B =Bx+By+Bz
where Ax, Ay, Az and Bx, By, and Bz are the calculated values of the x ‑, y ‑, and z ‑components of vectors A and B , respectively.

A = -14.87 i ^ + 8.42 j ^ + 0 k ^

B = -25.41 i ^ -12.0 j ^ + 0 k ^

Explanation:

For this exercise let's use trigonometry by decomposing to vectors

vector A

module 17.1 with an angle of 150.5 counterclockwise.

         Sin 150.5 = / A

         cos 150.5 = Ax / A

         A_{y} = A sin 150.5 = 17.1 sin 150.5

         Aₓ = A cos 1505 = 172 cos 150.5

         A_{y} = 8,420

         Aₓ = -14.870

the vector is

          A = -14.87 i ^ + 8.42 j ^ + 0 k ^

Vector B

          = 28.1 sin 205.3

         Bₓ = 28.1 cos 205.3

         B_{y} = -12.009

          Bₓ = -25.405

the vector is

          B = -25.41 i ^ -12.0 j ^ + 0 k ^

i think it’s a. this is because if they measured the mass of something with a mass of 150g, then their final answer would have to be approximately 150g. though all of their answers are almost 150, rosa and tran have the closest answers (or the exact answer) to the accurate one.

distance = 150m

displ = root (50 ^2 + 100^2)

root(2500+10000)

root 125000