The position vector for a proton is initially r(with vector sign above it) =5.0i(with top)- 6.0j(with top) + 2.0k(with top) and then later is r vector =
-2.0i(with top)+6.0j(with top)+ 2.0k (with top), all in meters. What is the proton's displacement vector, and to what plane is that vector parallel?
To find the proton's displacement vector, we subtract the initial position vector from the final position vector:
Displacement vector = final position vector - initial position vector
Given the initial position vector r₁ = 5.0î - 6.0ĵ + 2.0k̂ and the final position vector r₂ = -2.0î + 6.0ĵ + 2.0k̂, we can calculate the displacement vector as follows:
Δr = r₂ - r₁
= (-2.0î + 6.0ĵ + 2.0k̂) - (5.0î - 6.0ĵ + 2.0k̂)
= (-2.0î + 6.0ĵ + 2.0k̂) - 5.0î + 6.0ĵ - 2.0k̂
= -7.0î + 12.0ĵ
Therefore, the proton's displacement vector is -7.0î + 12.0ĵ.
To determine to which plane this vector is parallel, we look at the coefficients of the î, ĵ, and k̂ unit vectors. Since there is no component along the k̂ direction in the displacement vector, it is parallel to the xy-plane.
Hence, the proton's displacement vector is parallel to the xy-plane.