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Chapter 7: Problem 17

For any vector \(\overrightarrow{\mathbf{v}}=V_{x} \hat{\mathbf{i}}+V_{y} \hat{\mathbf{j}}+V_{z} \hat{\mathbf{k}}\) show that $$ V_{x}=\hat{\mathbf{i}} \cdot \overrightarrow{\mathbf{v}}, \quad V_{y}=\hat{\mathbf{j}} \cdot \overrightarrow{\mathbf{v}}, \quad V_{z}=\hat{\mathbf{k}} \cdot \overrightarrow{\mathbf{v}} $$

Short Answer

Expert verified
The vector components \( V_x \), \( V_y \), and \( V_z \) are given by the dot products \( \hat{\mathbf{i}} \cdot \overrightarrow{\mathbf{v}} \), \( \hat{\mathbf{j}} \cdot \overrightarrow{\mathbf{v}} \), and \( \hat{\mathbf{k}} \cdot \overrightarrow{\mathbf{v}} \), respectively.

Step by step solution

01

Identify the Vector Components

We are given a vector \( \overrightarrow{\mathbf{v}} = V_{x} \hat{\mathbf{i}} + V_{y} \hat{\mathbf{j}} + V_{z} \hat{\mathbf{k}} \). This vector is defined in terms of three components: \( V_{x} \), \( V_{y} \), and \( V_{z} \), along its unit vectors \( \hat{\mathbf{i}} \), \( \hat{\mathbf{j}} \), and \( \hat{\mathbf{k}} \). We need to show relationships involving dot products of these unit vectors with \( \overrightarrow{\mathbf{v}} \).
02

Dot Product with \( \hat{\mathbf{i}} \)

Compute the dot product of \( \overrightarrow{\mathbf{v}} \) with \( \hat{\mathbf{i}} \):\[\hat{\mathbf{i}} \cdot \overrightarrow{\mathbf{v}} = \hat{\mathbf{i}} \cdot (V_{x} \hat{\mathbf{i}} + V_{y} \hat{\mathbf{j}} + V_{z} \hat{\mathbf{k}})\]Using the distributive property of the dot product:\[= V_{x} (\hat{\mathbf{i}} \cdot \hat{\mathbf{i}}) + V_{y} (\hat{\mathbf{i}} \cdot \hat{\mathbf{j}}) + V_{z} (\hat{\mathbf{i}} \cdot \hat{\mathbf{k}})\]And knowing that \( \hat{\mathbf{i}} \cdot \hat{\mathbf{i}} = 1 \), \( \hat{\mathbf{i}} \cdot \hat{\mathbf{j}} = 0 \), \( \hat{\mathbf{i}} \cdot \hat{\mathbf{k}} = 0 \), the expression simplifies to:\[= V_{x} \]
03

Dot Product with \( \hat{\mathbf{j}} \)

Similarly, compute the dot product of \( \overrightarrow{\mathbf{v}} \) with \( \hat{\mathbf{j}} \):\[\hat{\mathbf{j}} \cdot \overrightarrow{\mathbf{v}} = \hat{\mathbf{j}} \cdot (V_{x} \hat{\mathbf{i}} + V_{y} \hat{\mathbf{j}} + V_{z} \hat{\mathbf{k}})\]Using the distributive property:\[= V_{x} (\hat{\mathbf{j}} \cdot \hat{\mathbf{i}}) + V_{y} (\hat{\mathbf{j}} \cdot \hat{\mathbf{j}}) + V_{z} (\hat{\mathbf{j}} \cdot \hat{\mathbf{k}})\]And knowing that \( \hat{\mathbf{j}} \cdot \hat{\mathbf{i}} = 0 \), \( \hat{\mathbf{j}} \cdot \hat{\mathbf{j}} = 1 \), \( \hat{\mathbf{j}} \cdot \hat{\mathbf{k}} = 0 \), the expression simplifies to:\[= V_{y} \]
04

Dot Product with \( \hat{\mathbf{k}} \)

Finally, compute the dot product of \( \overrightarrow{\mathbf{v}} \) with \( \hat{\mathbf{k}} \):\[\hat{\mathbf{k}} \cdot \overrightarrow{\mathbf{v}} = \hat{\mathbf{k}} \cdot (V_{x} \hat{\mathbf{i}} + V_{y} \hat{\mathbf{j}} + V_{z} \hat{\mathbf{k}})\]Using the distributive property:\[= V_{x} (\hat{\mathbf{k}} \cdot \hat{\mathbf{i}}) + V_{y} (\hat{\mathbf{k}} \cdot \hat{\mathbf{j}}) + V_{z} (\hat{\mathbf{k}} \cdot \hat{\mathbf{k}})\]And knowing that \( \hat{\mathbf{k}} \cdot \hat{\mathbf{i}} = 0 \), \( \hat{\mathbf{k}} \cdot \hat{\mathbf{j}} = 0 \), \( \hat{\mathbf{k}} \cdot \hat{\mathbf{k}} = 1 \), the expression simplifies to:\[= V_{z} \]
05

Conclusion of Dot Products

From the calculations, we have shown the relationships as:\[V_{x} = \hat{\mathbf{i}} \cdot \overrightarrow{\mathbf{v}}\]\[V_{y} = \hat{\mathbf{j}} \cdot \overrightarrow{\mathbf{v}}\]\[V_{z} = \hat{\mathbf{k}} \cdot \overrightarrow{\mathbf{v}}\]This verifies that the components of the vector match the dot product results of each unit vector with \( \overrightarrow{\mathbf{v}} \).

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Key Concepts

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

Dot Product
The dot product, often called the scalar product, is a mathematical operation that takes two equal-length sequences of numbers, usually coordinate vectors, and returns a single number. It combines components of the vectors in a way that
  • it results in a scalar (hence "scalar" product).
  • It reflects the product of the magnitudes of the two vectors and the cosine of the angle between them.
This makes the dot product a very useful tool in physics and engineering for determining how much of one's vector's direction aligns with another.
The general formula for two vectors \( \mathbf{A} \) and \( \mathbf{B} \) is given by \[ \mathbf{A} \cdot \mathbf{B} = A_{x}B_{x} + A_{y}B_{y} + A_{z}B_{z}\]where \( A_{x}, A_{y}, A_{z} \) and \( B_{x}, B_{y}, B_{z} \) are the components of vectors \({\mathbf{A}}\) and \({\mathbf{B}}\) respectively.

In this specific exercise, the dot product allows us to extract a specific component of the vector \( \overrightarrow{\mathbf{v}} \) by projecting it on the direction of the unit vectors \( \hat{\mathbf{i}}, \hat{\mathbf{j}}, \hat{\mathbf{k}} \). This operation pulls out the component that is aligned with the unit vector used in the calculation, reflecting the original statement we aim to demonstrate.
Unit Vectors
Unit vectors play a pivotal role in physics and vector calculations. A unit vector is a vector that has a magnitude of one and points in a specific direction. By using unit vectors,
  • We can define directions in three-dimensional space.
  • They have unique properties where the dot product between any two different unit vectors is zero. This is because they are perpendicular to each other.
The common unit vectors in the Cartesian coordinate system are \( \hat{\mathbf{i}}, \hat{\mathbf{j}}, \hat{\mathbf{k}} \), which point in the positive x, y, and z directions respectively.

These unit vectors are crucial when it comes to breaking down other vectors into their components. In our exercise, they serve to extract \( V_x, V_y, \) and \( V_z \) from \( \overrightarrow{\mathbf{v}} \). The process involves a simple dot product operation as each unit vector pulls out only the component of \( \overrightarrow{\mathbf{v}} \) that lies in the same direction, utilizing the rules of orthogonality and normalization inherent to unit vectors.
Physics Education
Physics education often uses vectors as a foundational concept because they describe quantities with both magnitude and direction. Understanding vectors allows students to better understand a variety of physical phenomena from basic motion to complex forces.
  • Vectors help represent vast areas of physics like velocity, acceleration, and forces.
  • Learning the mathematics of vectors, such as dot products and unit vectors, strengthens the conceptual understanding of students.
Teaching techniques around vectors should focus on visualization and real-world analogies to help students connect the mathematical formalism with the physical concepts.

Exercises such as the one provided reinforce the idea that vectors can be broken down into simpler parts, making complex problems more manageable. Breaking vectors into components using unit vectors and dot products emphasizes how each direction can contribute independently to the entire vector's behavior.
This kind of step-by-step breakdown not only verifies mathematical operations but also demonstrates the practical aspect of vectors in solving physics problems effectively.

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Most popular questions from this chapter

A \(75.0-\mathrm{kg}\) firefighter climbs a flight of stairs \(20.0 \mathrm{~m}\) high. How much work is required?

How much work is required to stop an electron \(\left(m=9.11 \times 10^{-31} \mathrm{~kg}\right)\) which is moving with a speed of \(1.40 \times 10^{6} \mathrm{~m} / \mathrm{s} ?\)

\(\begin{array}{llll}\text { (II) } & \text { Suppose } & \mathbf{A}=1.0 \mathbf{i}+1.0 \mathbf{j}-2.0 \mathbf{k} & \text { and } & \mathbf{B}=\end{array}\) \(-1.0 \hat{\mathrm{i}}+1.0 \mathrm{j}+2.0 \hat{\mathbf{k}}, \quad(a)\) what is the angle between these two vectors? \((b)\) Explain the significance of the sign in part \((a)\).

An elevator cable breaks when a \(925-\mathrm{kg}\) elevator is \(22.5 \mathrm{~m}\) above the top of a huge spring \((k=\) \(\left.8.00 \times 10^{4} \mathrm{~N} / \mathrm{m}\right)\) at the bottom of the shaft. Calculate (a) the work done by gravity on the elevator before it hits the spring; (b) the speed of the elevator just before striking the spring; \((c)\) the amount the spring compresses (note that here work is done by both the spring and gravity).

A 3.0 -m-long steel chain is stretched out along the top level of a horizontal scaffold at a construction site, in such a way that \(2.0 \mathrm{~m}\) of the chain remains on the top level and \(1.0 \mathrm{~m}\) hangs vertically, Fig. \(7-26 .\) At this point, the force on the hanging segment is sufficient to pull the entire chain over the edge. Once the chain is moving, the kinetic friction is so small that it can be neglected. How much work is performed on the chain by the force of gravity as the chain falls from the point where \(2.0 \mathrm{~m}\) remains on the scaffold to the point where the entire chain has left the scaffold? (Assume that the chain has a linear weight density of \(18 \mathrm{~N} / \mathrm{m}\).)
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