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Chapter 14: Problem 5

Sketch the position vector of a and find \(\|\mathrm{a}\|\). $$ a=-4 i+5 j $$

Short Answer

Expert verified
The magnitude of vector \( \mathbf{a} = -4 \mathbf{i} + 5 \mathbf{j} \) is \( \sqrt{41} \).

Step by step solution

01

Interpret the Position Vector

The vector \( \mathbf{a} \) is given as \( -4 \mathbf{i} + 5 \mathbf{j} \). This represents a vector in a 2D plane, where \( \mathbf{i} \) is the unit vector along the x-axis and \( \mathbf{j} \) is the unit vector along the y-axis. Therefore, \( \mathbf{a} \) can be seen as pointing from the origin \((0, 0)\) to the point \((-4, 5)\).
02

Sketch the Vector

To sketch the vector \( \mathbf{a} \), start at the origin \((0, 0)\) on a graph. From there, move 4 units to the left (since the x-component is -4) and 5 units up (since the y-component is 5). Plot the point \((-4, 5)\) and draw an arrow from the origin to this point to represent the vector.
03

Determine the Magnitude of the Vector

The magnitude of a vector \( \mathbf{a} = a_x \mathbf{i} + a_y \mathbf{j} \) is calculated using the formula: \[ \| \mathbf{a} \| = \sqrt{a_x^2 + a_y^2} \] For the vector \( \mathbf{a} = -4 \mathbf{i} + 5 \mathbf{j} \), we have \( a_x = -4 \) and \( a_y = 5 \). Substituting these values in, we get: \[ \| \mathbf{a} \| = \sqrt{(-4)^2 + 5^2} = \sqrt{16 + 25} = \sqrt{41} \]
04

Finalize the Magnitude Calculation

After computing \( \sqrt{41} \), the magnitude of the vector is left as is because \( \sqrt{41} \) cannot be simplified further without approximation. Therefore, the magnitude of vector \( \mathbf{a} \) is \( \sqrt{41} \).

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Key Concepts

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

Magnitude of a Vector
The magnitude of a vector is essential for understanding its length or size in the space where it exists. For any vector represented in the form \( \mathbf{a} = a_x \mathbf{i} + a_y \mathbf{j} \), its magnitude can be calculated using the Pythagorean theorem:
  • Square each component of the vector.
  • Add the results of these squares.
  • Take the square root of the sum.
This gives the formula: \[\| \mathbf{a} \| = \sqrt{a_x^2 + a_y^2}\]In the case of the vector \( \mathbf{a} = -4 \mathbf{i} + 5 \mathbf{j} \), we follow this process:
  • The \( x \)-component is \(-4\), and the \( y \)-component is \(5\).
  • Calculate \((-4)^2 = 16\) and \(5^2 = 25\).
  • Add these results to get \(41\).
  • Finally, take the square root to find \( \sqrt{41} \), the vector's magnitude.
This helps us comprehend the actual "length" of the vector when drawn from the origin to its terminal point.
2D Plane Vectors
When dealing with vectors, it's crucial to understand the foundational concept of vectors existing in a 2D plane. A 2D plane is simply a flat surface that extends infinitely in two directions, typically defined by the x-axis and y-axis. Vectors in this space have two components:
  • An x-component which lies along the horizontal axis.
  • A y-component which lies along the vertical axis.
Thus, any 2D vector like \( \mathbf{a} = -4 \mathbf{i} + 5 \mathbf{j} \) can be seen as an instruction for movement within this plane:
  • The \(-4 \mathbf{i}\) means moving 4 units to the left along the x-direction.
  • The "+5 \mathbf{j}" suggests climbing 5 units upward along the y-direction.
Every vector on a 2D plane can be expressed as a combination of these two components, which helps in calculations like finding its magnitude, direction, and usefulness in real-world applications such as physics and engineering.
Unit Vectors
Unit vectors are the building blocks of vector mathematics in 2D and 3D geometry. They represent direction but have a magnitude of exactly one. In a 2D plane:
  • The unit vector \( \mathbf{i} \) points along the x-axis and has a magnitude of 1 unit.
  • The unit vector \( \mathbf{j} \) aligns along the y-axis and also has a magnitude of 1 unit.
These vectors allow us to express other vectors in terms of pure direction on a coordinate system without affecting their size.When any vector, such as \( \mathbf{a} = -4 \mathbf{i} + 5 \mathbf{j} \), is represented, it is typically broken down into its unit vector components to clarify where it is pointing in space:
  • \(-4 \mathbf{i} \) indicates movement in the negative x direction.
  • \(5 \mathbf{j} \) shows movement in the positive y direction.
Understanding unit vectors simplifies complex vector operations because it allows dimensions of larger vectors to be directly compared or added without transforming their intrinsic properties.
Vector Sketching
Sketching a vector allows us to visually appreciate its properties and direction in a 2D plane. It combines understanding its magnitude and components:
  • Vectors are typically sketched by drawing an arrow starting from the origin \((0, 0)\).
  • Move the number of units equal to the x-component along the x-axis.
  • From there, move the units indicated by the y-component along the y-axis.
  • Finally, draw an arrow from the origin to the endpoint you arrived at.
For example, sketching the vector \( \mathbf{a} = -4 \mathbf{i} + 5 \mathbf{j} \) involves:
  • Starting at the origin.
  • Moving 4 units left due to \(-4 \mathbf{i}\).
  • Then climbing up 5 units because of \(5 \mathbf{j}\).
  • The endpoint of this movement, \((-4, 5)\), is where you draw the vector's head.
This simple yet effective graphical representation helps in analyzing vector behavior, interaction with other vectors, and its application in various domains.

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