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Found in: Page 91

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# The position vector for a proton is initially $\stackrel{\mathbf{â†’}}{\mathbf{r}}{\mathbf{=}}{\mathbf{5}}{\mathbf{.}}{\mathbf{0}}\stackrel{\mathbf{^}}{\mathbf{i}}{\mathbf{-}}{\mathbf{6}}{\mathbf{.}}{\mathbf{0}}\stackrel{\mathbf{^}}{\mathbf{j}}{\mathbf{+}}{\mathbf{2}}{\mathbf{.}}{\mathbf{0}}\stackrel{\mathbf{^}}{\mathbf{k}}$ and then later is role="math" localid="1657003791208" $\stackrel{\mathbf{â†’}}{\mathbf{r}}{\mathbf{=}}{\mathbf{-}}{\mathbf{2}}{\mathbf{.}}{\mathbf{0}}\stackrel{\mathbf{^}}{\mathbf{i}}{\mathbf{-}}{\mathbf{6}}{\mathbf{.}}{\mathbf{0}}\stackrel{\mathbf{^}}{\mathbf{j}}{\mathbf{+}}{\mathbf{2}}{\mathbf{.}}{\mathbf{0}}\stackrel{\mathbf{^}}{\mathbf{k}}$ all in meters. (a) What is the protonâ€™s displacement vector, and (b) to what plane is that vector parallel?

(a) The protonâ€™s displacement vector is$\left(-7\stackrel{^}{i}+12\stackrel{^}{j}\right)m$

(b) $\stackrel{â†’}{d}$is parallel to the X-Y plane

See the step by step solution

## Step 1: Given data

1) Initial position vector ${\stackrel{â†’}{\mathrm{r}}}_{\mathrm{i}}=5.0\stackrel{^}{\mathrm{i}}-6.0\stackrel{^}{\mathrm{j}}+2.0\stackrel{^}{\mathrm{k}}$

2) Final position vector $\stackrel{â†’}{{\mathrm{r}}_{\mathrm{f}}}=-2.0\stackrel{^}{\mathrm{i}}-6.0\stackrel{^}{\mathrm{j}}+2.0\stackrel{^}{\mathrm{k}}$

## Step 2: Understanding the concept of vector addition and subtraction

Taking the difference between position vectors of proton, we can find the displacement vector. We have to use vector laws of addition for this.

Formula:

$\stackrel{â†’}{d}={\stackrel{â†’}{r}}_{f}-{\stackrel{â†’}{r}}_{i}$

## Step 3: (a) Determining the displacement vector

To determine the displacement$\stackrel{â†’}{d}$, we need to use the equation,

$\stackrel{â†’}{d}={\stackrel{â†’}{r}}_{f}-{\stackrel{â†’}{r}}_{i}\phantom{\rule{0ex}{0ex}}=\left(-2.0\stackrel{^}{i}+6.0\stackrel{^}{j}+2.0\stackrel{^}{k}\right)m-\left(5.0\stackrel{^}{i}-6.0\stackrel{^}{j}+2.0\stackrel{^}{k}\right)m\phantom{\rule{0ex}{0ex}}=\left(-7.0\stackrel{^}{i}+12\stackrel{^}{j}\right)m$

Therefore, the displacement vector is $\left(-7.0\stackrel{^}{i}+12\stackrel{^}{j}\right)m$

## Step 4: (b) Determining the direction of the plane

The displacement vector $\stackrel{â†’}{d}$ does not have any component along Z-axis. It means that the displacement vector lies in XY plane.Therefore, the displacement vector is parallel to the XY plane.