91Ó°ÊÓ

-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. -2 A watermelon seed has the following coordinates: \(x=-5.0 \mathrm{~m}\), \(y=8.0 \mathrm{~m}\), and \(z=0 \mathrm{~m} .\) Find its position vector (a) in unit-vector notation and as (b) a magnitude and (c) an angle relative to the positive direction of the \(x\) axis. (d) Sketch the vector on a right-handed coordinate system. If the seed is moved to the \(x y z\) coordinates \((3.00 \mathrm{~m},\), \(0 \mathrm{~m}, 0 \mathrm{~m}\) ), what is its displacement (e) in unit-vector notation and as (f) a magnitude and (g) an angle relative to the positive \(x\) direction?

Step by step solution

01

Magnitude of Vector \( \vec{r} \)

The magnitude of a vector \( \vec{r} = (x, y, z) \) is calculated by applying the formula: \[ |\vec{r}| = \sqrt{x^2 + y^2 + z^2} \]. For \( \vec{r} = (5.0\, \mathrm{m}) \hat{\mathrm{i}} - (3.0\, \mathrm{m}) \hat{\mathrm{j}} + (2.0\, \mathrm{m}) \hat{\mathrm{k}} \), substitute the components into the formula: \[ |\vec{r}| = \sqrt{(5.0)^2 + (-3.0)^2 + (2.0)^2} = \sqrt{25 + 9 + 4} = \sqrt{38} \]. This simplifies to approximately \( 6.16 \, \mathrm{m} \).

02

Sketching the Vector \( \vec{r} \)

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 three-dimensional right-handed coordinate system. Begin at the origin. Move 5.0 meters in the positive \( x \)-direction, 3.0 meters in the negative \( y \)-direction, and then 2.0 meters in the positive \( z \)-direction. Mark the terminal point of this path and draw the vector \( \vec{r} \) from the origin to this point.

03

Position Vector of Watermelon Seed in Unit-Vector Notation

For a watermelon seed with coordinates \( x = -5.0\, \mathrm{m} \), \( y = 8.0\, \mathrm{m} \), and \( z = 0 \), express its position vector in unit-vector notation as: \( \vec{r} = (-5.0\, \mathrm{m}) \hat{\mathrm{i}} + (8.0\, \mathrm{m}) \hat{\mathrm{j}} \).

04

Magnitude of Watermelon Seed's Position Vector

Using similar magnitude calculation, \( |\vec{r}| = \sqrt{(-5.0)^2 + (8.0)^2 + (0)^2} = \sqrt{25 + 64} = \sqrt{89} \), which simplifies to approximately \( 9.43 \, \mathrm{m} \).

05

Angle Relative to the Positive X-axis

To find the angle \( \theta \) of the vector relative to the positive \( x \)-axis, use \( \tan \theta = \frac{y}{x} = \frac{8.0}{-5.0} \). Calculate \( \theta \) using \( \theta = \arctan\left(\frac{8.0}{-5.0}\right) \), which gives an angle of approximately \(-58.0^\circ \). Adjust to positive angles by adding \(180^\circ\) to get \(122.0^\circ\).

06

Sketching the Watermelon Seed's Position Vector

In the same coordinate system, plot a point at \( (-5.0\, \mathrm{m}, 8.0\, \mathrm{m}, 0) \) and draw the vector \( \vec{r} \) from the origin to this point.

07

Calculating the Displacement in Unit-Vector Notation

For the displacement from the original position \((-5.0\, \mathrm{m}, 8.0\, \mathrm{m}, 0)\) to the new position \((3.0\, \mathrm{m}, 0, 0)\), compute the difference: \[\vec{d} = (3.0 - (-5.0)\, \mathrm{m})\hat{\mathrm{i}} + (0 - 8.0\, \mathrm{m})\hat{\mathrm{j}} + (0 - 0)\, \hat{\mathrm{k}} = (8.0\, \mathrm{m})\hat{\mathrm{i}} + (-8.0\, \mathrm{m})\hat{\mathrm{j}} \].

08

Magnitude of Displacement

Find the magnitude of this displacement vector using: \[ |\vec{d}| = \sqrt{(8.0)^2 + (-8.0)^2} = \sqrt{64 + 64} = \sqrt{128} \], which simplifies to approximately \( 11.31 \, \mathrm{m} \).

09

Angle of Displacement Relative to Positive X-axis

Calculate the angle of the displacement vector \( \vec{d} \) using \( \tan\theta = \frac{-8.0}{8.0} = -1 \). Thus, \( \theta = 135^\circ \) since the vector lies in the second quadrant.

10

Sketching the Displacement Vector

Draw on the coordinate system a vector \( \vec{d} \) from the original position to the new position at \( (3.0\, \mathrm{m},0,0) \). The vector should represent a movement of 8 meters in both \( x \) and negative \( y \) directions.

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Key Concepts

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

Position Vector

In vector calculus, a position vector is a crucial concept used to specify the location of a point in space relative to a chosen origin. It is represented as a vector originating from the origin to the point in question. In the case of a watermelon seed given coordinates

• \( x = -5.0 \, \mathrm{m} \)
• \( y = 8.0 \, \mathrm{m} \)
• \( z = 0 \, \mathrm{m} \)

Its position vector in unit-vector notation is expressed as \( \vec{r} = (-5.0 \, \mathrm{m}) \hat{\mathrm{i}} + (8.0 \, \mathrm{m}) \hat{\mathrm{j}} \).
This unit-vector notation concisely conveys both the direction and magnitude or distance from the origin to the point along each coordinate axis. Here, the negative sign of the \( i \)-component indicates movement in the negative \( x \)-axis direction.
Understanding position vectors is essential as they describe the initial setup for performing operations such as finding magnitudes and calculating angles.

Magnitude of Vector

The magnitude of a vector represents its length in space. Calculating the magnitude of a vector is significant as it provides the scalar measure of the vector irrespective of its components' direction. The formula to determine the magnitude of a vector \( \vec{r} = (x, y, z) \) is:

• \[ |\vec{r}| = \sqrt{x^2 + y^2 + z^2} \]

For our watermelon seed position vector \( \vec{r} = (-5.0 \, \mathrm{m}) \hat{\mathrm{i}} + (8.0 \, \mathrm{m}) \hat{\mathrm{j}} \), albeit the seed's \( z \) component being zero, we can still compute the magnitude using:

• \[ |\vec{r}| = \sqrt{(-5.0)^2 + (8.0)^2 + (0)^2} = \sqrt{25 + 64} = \sqrt{89} \]
• \( |\vec{r}| \approx 9.43 \, \mathrm{m} \)

Magnitudes are always positive, representing distance in the geometrical sense. This 9.43 meters reveals how far the seed's point is from the origin along a direct path.

Angle Relative to Axis

Angles between vectors and the coordinate axes are essential to understand the orientation of vectors in space. Here, we explore the angle between a position vector and the positive \( x \)-axis. This involves using trigonometry, specifically the tangent function, which relates the angle to the vector's components. The formula is

• \[ \tan \theta = \frac{y}{x} \]

For our position vector: - With \( x = -5.0 \, \mathrm{m} \) and \( y = 8.0 \, \mathrm{m} \), - The angle \( \theta \) with respect to the positive \( x \)-axis is calculated by

• \[ \theta = \arctan\left(\frac{8.0}{-5.0}\right) \]

This yields approximately

• \( \theta = -58.0^\circ \).
• Adjusted to a positive angle in standard position (from the positive \( x \)-axis counterclockwise) by adding \(180^\circ\).
• Resulting in approximately \( 122.0^\circ \).

This angle's computation aids in visually and analytically understanding the directional relationship of vectors within the coordinate system.