/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 106 The position vector for a proton... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \(\mathrm{in}\) meters. (a) What is the proton's displacement vector, and (b) to what plane is that vector parallel?

Short Answer

Expert verified
(a) \( \vec{d} = -7.0 \hat{\mathrm{i}} + 12.0 \hat{\mathrm{j}} \); (b) parallel to the \( xy \)-plane.

Step by step solution

01

Understand the Initial and Final Position Vectors

The initial position vector \( \vec{r}_i \) is given as \( \vec{r}_i = 5.0 \hat{\mathrm{i}} - 6.0 \hat{\mathrm{j}} + 2.0 \hat{\mathrm{k}} \). The final position vector \( \vec{r}_f \) is \( \vec{r}_f = -2.0 \hat{\mathrm{i}} + 6.0 \hat{\mathrm{j}} + 2.0 \hat{\mathrm{k}} \). Both vectors are in meters.
02

Calculate the Displacement Vector

The displacement vector \( \vec{d} \) is found by subtracting the initial position vector from the final position vector: \( \vec{d} = \vec{r}_f - \vec{r}_i \). This yields \( \vec{d} = (-2.0 \hat{\mathrm{i}} + 6.0 \hat{\mathrm{j}} + 2.0 \hat{\mathrm{k}}) - (5.0 \hat{\mathrm{i}} - 6.0 \hat{\mathrm{j}} + 2.0 \hat{\mathrm{k}}) \).
03

Perform the Vector Subtraction

Carry out the subtraction component-wise: \(-2.0 - 5.0 = -7.0\) for \( \hat{\mathrm{i}} \), \(6.0 - (-6.0) = 12.0\) for \( \hat{\mathrm{j}} \), and \(2.0 - 2.0 = 0.0\) for \( \hat{\mathrm{k}} \). Therefore, the displacement vector \( \vec{d} \) is \(-7.0 \hat{\mathrm{i}} + 12.0 \hat{\mathrm{j}} + 0.0 \hat{\mathrm{k}} \).
04

Identify the Plane the Vector is Parallel To

The vector \( -7.0 \hat{\mathrm{i}} + 12.0 \hat{\mathrm{j}} + 0.0 \hat{\mathrm{k}} \) has no component in the \( \hat{\mathrm{k}} \)-direction, indicating it lies in the \( xy \)-plane. Thus, the displacement vector is parallel to the \( xy \)-plane.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Displacement Vector
A displacement vector signifies a change in position of an object in space. For any moving object, the displacement vector connects its initial position to its final position. In this example, a proton has an initial position vector \( \vec{r}_i = 5.0 \hat{\mathrm{i}} - 6.0 \hat{\mathrm{j}} + 2.0 \hat{\mathrm{k}} \) and a final position vector \( \vec{r}_f = -2.0 \hat{\mathrm{i}} + 6.0 \hat{\mathrm{j}} + 2.0 \hat{\mathrm{k}} \). This simply tells us where the proton started and where it ended.
The displacement vector, \( \vec{d} \), is calculated by subtracting the initial position vector from the final position vector to find the change in position. This is done with the formula:\( \vec{d} = \vec{r}_f - \vec{r}_i \). This operation will help us understand not only how far the proton has traveled but also in which directions it has moved between its initial and final states.
Vector Subtraction
Vector subtraction is a method used to find the difference between two vectors, often employed to ascertain displacement like in our exercise. The principle involves a component-wise subtraction of one vector from another. For our proton, the displacement vector is found by subtracting each component of the initial position from the corresponding component of the final position. Let's break it down:
  • For the \( \hat{\mathrm{i}} \) component: \( -2.0 - 5.0 = -7.0 \)
  • For the \( \hat{\mathrm{j}} \) component: \( 6.0 - (-6.0) = 12.0 \)
  • For the \( \hat{\mathrm{k}} \) component: \( 2.0 - 2.0 = 0.0 \)

This gives us a displacement vector \( \vec{d} = -7.0 \hat{\mathrm{i}} + 12.0 \hat{\mathrm{j}} + 0.0 \hat{\mathrm{k}} \). Thanks to vector subtraction, we can tell that the proton has moved 7 units left and 12 units upward, with no change in the \( \hat{\mathrm{k}} \) direction.
Coordinate Systems
Coordinate systems are frameworks that allow us to uniquely identify every point in space using numbers, known as vector components. In physics, particularly when dealing with vectors, the Cartesian coordinate system (composed of \( \hat{\mathrm{i}}, \hat{\mathrm{j}}, \hat{\mathrm{k}} \) unit vectors) is most frequently used. Each of these components represents movement along one of the three principal axes – \( x, y, \) and \( z \), respectively.
For the proton's position:
  • The \( \hat{\mathrm{i}} \) component indicates movement along the x-axis.
  • The \( \hat{\mathrm{j}} \) component refers to movement along the y-axis.
  • The \( \hat{\mathrm{k}} \) component describes movement along the z-axis.

In the exercise at hand, the displacement vector \( -7.0 \hat{\mathrm{i}} + 12.0 \hat{\mathrm{j}} + 0.0 \hat{\mathrm{k}} \) suggests parallelism with the \( xy \)-plane because it has no \( \hat{\mathrm{k}} \) component. Thus, understanding coordinate systems helps us visualize our vector in three-dimensional space and quickly determine its orientation or direction relative to the axes and planes.

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