Solution Found!
The position vector for a proton is initially 5.0i 6.0j 2.0k r: and then later is 2.0i
Chapter 4, Problem 106(choose chapter or problem)
The position vector for a proton is initially \(\vec{r}\) = \(5.0 \hat{\mathrm{i}}-6.0 \hat{\mathrm{j}}+2.0 \hat{\mathrm{k}}\) and then later is \(\vec{r}=-2.0 \hat{\mathrm{i}}+6.0 \hat{\mathrm{j}}+2.0 \hat{\mathrm{k}}\), all in meters.
(a) What is the proton’s displacement vector, and
(b) to what plane is that vector parallel?
Questions & Answers
QUESTION:
The position vector for a proton is initially \(\vec{r}\) = \(5.0 \hat{\mathrm{i}}-6.0 \hat{\mathrm{j}}+2.0 \hat{\mathrm{k}}\) and then later is \(\vec{r}=-2.0 \hat{\mathrm{i}}+6.0 \hat{\mathrm{j}}+2.0 \hat{\mathrm{k}}\), all in meters.
(a) What is the proton’s displacement vector, and
(b) to what plane is that vector parallel?
ANSWER:Step 1 of 3
Part (a)
The protons displacement vector can be determined by taking the difference between initial and final position vectors.
The protons displacement vector can be written as,
\(\Delta \vec{r}=\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\)