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Chapter 4: Problem 1

The position vector for an electron is \(\vec{r}=(5.0 \mathrm{~m}) \hat{\mathrm{i}}-\) \((3.0 \mathrm{~m}) \hat{\mathrm{j}}+(2.0 \mathrm{~m}) \hat{\mathrm{k}}\). (a) Find the magnitude of \(\vec{r}\). (b) Sketch the vector on a right-handed coordinate system.

Short Answer

Expert verified
(a) Magnitude is approximately 6.16 meters. (b) Sketch on 3D axes pointing to (5.0, -3.0, 2.0).

Step by step solution

01

Identify the Position Vector

The given position vector is \( \vec{r} = (5.0 \mathrm{~m}) \hat{\mathrm{i}} - (3.0 \mathrm{~m}) \hat{\mathrm{j}} + (2.0 \mathrm{~m}) \hat{\mathrm{k}} \). This vector can be broken down into its components: \( x = 5.0 \mathrm{~m} \), \( y = -3.0 \mathrm{~m} \), and \( z = 2.0 \mathrm{~m} \).
02

Calculate Magnitude of the Vector

The magnitude of a vector \( \vec{r} = x \hat{\mathrm{i}} + y \hat{\mathrm{j}} + z \hat{\mathrm{k}} \) is given by the formula \( ||\vec{r}|| = \sqrt{x^2 + y^2 + z^2} \). Substitute the values: \( ||\vec{r}|| = \sqrt{(5.0)^2 + (-3.0)^2 + (2.0)^2} \). Simplifying gives \( ||\vec{r}|| = \sqrt{25 + 9 + 4} = \sqrt{38} \approx 6.16 \mathrm{~m} \).
03

Sketching the Vector

To sketch the vector \( \vec{r} = (5.0 \mathrm{~m}) \hat{\mathrm{i}} - (3.0 \mathrm{~m}) \hat{\mathrm{j}} + (2.0 \mathrm{~m}) \hat{\mathrm{k}} \), draw a 3D coordinate system with axes labeled \( x \), \( y \), and \( z \). The vector begins at the origin and extends to the point \( (5.0, -3.0, 2.0) \), where these values correspond to the coordinates along the \( x \), \( y \), and \( z \) axes, respectively.

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Key Concepts

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

Position Vector
A position vector is essential in describing where a point is located in a 3D space. It represents the position of a point relative to the origin of the coordinate system. Imagine it as an arrow starting from the origin and ending at the given point in space.
In our exercise, the position vector is given as:
  • \( \vec{r} = (5.0 \mathrm{~m}) \hat{\mathrm{i}} - (3.0 \mathrm{~m}) \hat{\mathrm{j}} + (2.0 \mathrm{~m}) \hat{\mathrm{k}} \)
  • It consists of three components, with each alphabet (\( \hat{\mathrm{i}} \), \( \hat{\mathrm{j}} \), and \( \hat{\mathrm{k}} \)) representing one of the spatial directions: x, y, and z, respectively.
Each component tells you how far you move along a particular axis:
  • The value \( 5.0 \mathrm{~m} \) indicates the movement along the x-axis.
  • The value \( -3.0 \mathrm{~m} \) indicates the movement backward along the y-axis.
  • The value \( 2.0 \mathrm{~m} \) indicates the movement along the z-axis.
By comprehending these components, you can visualize the vector and understand how position vectors work in geometry and physics.
Magnitude of a Vector
The magnitude of a vector acts as a measure of its length. It tells us how long the vector is, without considering its direction. To compute the magnitude, especially in three-dimensional space, you can apply the Pythagorean theorem.
For a vector \( \vec{v} = x \hat{\mathrm{i}} + y \hat{\mathrm{j}} + z \hat{\mathrm{k}} \), the formula for its magnitude is:
  • \( ||\vec{v}|| = \sqrt{x^2 + y^2 + z^2} \)
Applying this to our position vector:
  • Substitute the values into the formula: \( \sqrt{(5.0)^2 + (-3.0)^2 + (2.0)^2} \)
  • This simplifies to \( \sqrt{25 + 9 + 4} = \sqrt{38} \)
  • The approximate magnitude is \( 6.16 \mathrm{~m} \)
This magnitude represents the straight-line distance from the origin to the point (5.0, -3.0, 2.0) in space. Understanding how to find the magnitude is vital in vector calculus, as it aids in determining distances and scalars.
3D Coordinate System
A 3D coordinate system is a framework that allows us to locate points in three-dimensional space. It uses three axes, typically named x, y, and z. These axes are perpendicular to each other and intersect at a point called the origin.
  • The x-axis is often drawn horizontally and usually points to the right.
  • The y-axis is typically drawn vertically and usually points up or backward.
  • The z-axis is usually drawn out of or into the page/screen, representing depth.
This system helps us visualize vectors and other geometric objects in space. In our exercise, the vector \( \vec{r} \) can be represented on this 3D plane:
  • It starts from the origin (0, 0, 0).
  • Ends at the coordinates \( (5.0, -3.0, 2.0) \).
Sketching vectors in a 3D coordinate system can significantly enhance our understanding of spatial relationships in fields like physics and engineering. Remember to always visualize how positive and negative values move us along and against the axes.

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Most popular questions from this chapter

A woman who can row a boat at \(6.4 \mathrm{~km} / \mathrm{h}\) in still water faces a long, straight river with a width of \(6.4 \mathrm{~km}\) and a current of \(3.2 \mathrm{~km} / \mathrm{h}\). Let \(\hat{\text { i }}\) point directly across the river and \(\hat{\mathrm{j}}\) point directly downstream. If she rows in a straight line to a point directly opposite her starting position, (a) at what angle to 卯 must she point the boat and (b) how long will she take? (c) How long will she take if, instead, she rows \(3.2 \mathrm{~km}\) down the river and then back to her starting point? (d) How long if she rows \(3.2 \mathrm{~km} u p\) the river and then back to her starting point? (e) At what angle to \(\hat{\mathbf{i}}\) should she point the boat if she wants to cross the river in the shortest possible time? (f) How long is that shortest time?

Snow is falling vertically at a constant speed of \(8.0 \mathrm{~m} / \mathrm{s}\). At what angle from the vertical do the snowflakes appear to be falling as viewed by the driver of a car traveling on a straight, level road with a speed of \(50 \mathrm{~km} / \mathrm{h} ?\)

Two seconds after being projected from ground level, a projectile is displaced \(40 \mathrm{~m}\) horizontally and \(53 \mathrm{~m}\) vertically above its launch point. What are the (a) horizontal and (b) vertical components of the initial velocity of the projectile? (c) At the instant the projectile achieves its maximum height above ground level, how far is it displaced horizontally from the launch point?

A particle moves along a circular path over a horizontal \(x y\) coordinate system, at constant speed. At time \(t_{1}=4.00 \mathrm{~s}\), it is at point \((5.00 \mathrm{~m}, 6.00 \mathrm{~m})\) with velocity \((3.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\) and acceleration in the positive \(x\) direction. At time \(t_{2}=10.0 \mathrm{~s}\), it has velocity \((-3.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}\) and acceleration in the positive \(y\) direction. What are the (a) \(x\) and (b) \(y\) coordinates of the center of the circular path if \(t_{2}-t_{1}\) is less than one period?

Two ships, \(A\) and \(B\), leave port at the same time. Ship \(A\) travels northwest at 24 knots, and ship \(B\) travels at 28 knots in a direction \(40^{\circ}\) west of south. \((1 \mathrm{knot}=1\) nautical mile per hour; see Appendix D.) What are the (a) magnitude and (b) direction of the velocity of ship \(A\) relative to \(B ?\) (c) After what time will the ships be 160 nautical miles apart? (d) What will be the bearing of \(B\) (the direction of \(B\) 's position) relative to \(A\) at that time?
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