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Given two vectors \(\overrightarrow{A}\) = \(-\)2.00\(\hat{\imath}\) \(+\) 3.00\(\hat{\jmath}\) \(+\) 4.00\(\hat{k}\) and \(\overrightarrow{B}\) = 3.00\(\hat{\imath}\) \(+\) 1.00\(\hat{\jmath}\) \(-\) 3.00\(\hat{k}\), (a) find the magnitude of each vector; (b) use unit vectors to write an expression for the vector difference \(\overrightarrow{A}\) \(-\) \(\overrightarrow{A}\); and (c) find the magnitude of the vector difference \(\overrightarrow{A}\) \(-\) \(\overrightarrow{A}\). Is this the same as the magnitude of \(\overrightarrow{A}\) \(-\) \(\overrightarrow{A}\)? Explain.

Short Answer

Expert verified
(a) Magnitude of A: 5.39, B: 4.36. (b) A - B = -5.00\hat{\imath} + 2.00\hat{\jmath} + 7.00\hat{k}. (c) Magnitude of A - B: 8.83. A - A is zero.

Step by step solution

01

Find the Magnitude of Vector A

To find the magnitude of vector \( \overrightarrow{A} = -2.00\hat{\imath} + 3.00\hat{\jmath} + 4.00\hat{k} \), use the formula: \( |\overrightarrow{A}| = \sqrt{(-2.00)^2 + (3.00)^2 + (4.00)^2} \). Calculate: \[ |\overrightarrow{A}| = \sqrt{4.00 + 9.00 + 16.00} = \sqrt{29.00} \approx 5.39 \].
02

Find the Magnitude of Vector B

For vector \( \overrightarrow{B} = 3.00\hat{\imath} + 1.00\hat{\jmath} - 3.00\hat{k} \), use the magnitude formula: \( |\overrightarrow{B}| = \sqrt{(3.00)^2 + (1.00)^2 + (-3.00)^2} \). This gives: \[ |\overrightarrow{B}| = \sqrt{9.00 + 1.00 + 9.00} = \sqrt{19.00} \approx 4.36 \].
03

Express the Vector Difference A - B

Subtract the components of \( \overrightarrow{B} \) from \( \overrightarrow{A} \) to get \( \overrightarrow{A} - \overrightarrow{B} = (-2.00 - 3.00)\hat{\imath} + (3.00 - 1.00)\hat{\jmath} + (4.00 + 3.00)\hat{k} \). Simplify this to: \( \overrightarrow{A} - \overrightarrow{B} = -5.00\hat{\imath} + 2.00\hat{\jmath} + 7.00\hat{k} \).
04

Find the Magnitude of Vector A - B

Calculate the magnitude of \( \overrightarrow{A} - \overrightarrow{B} = -5.00\hat{\imath} + 2.00\hat{\jmath} + 7.00\hat{k} \) using the formula: \( |\overrightarrow{A} - \overrightarrow{B}| = \sqrt{(-5.00)^2 + (2.00)^2 + (7.00)^2} \). Compute: \[ |\overrightarrow{A} - \overrightarrow{B}| = \sqrt{25.00 + 4.00 + 49.00} = \sqrt{78.00} \approx 8.83 \].
05

Compare Magnitude A - A and A - B

The vector \( \overrightarrow{A} - \overrightarrow{A} \) is the zero vector with a magnitude of 0, because subtracting a vector from itself results in the zero vector. Thus, \( |\overrightarrow{A} - \overrightarrow{A}| = 0 \). The magnitude of \( \overrightarrow{A} - \overrightarrow{B} \) is \( \approx 8.83 \), which is not the same as 0, confirming the results are different.

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Key Concepts

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

Vector Magnitude
The magnitude of a vector is a measure of its length. It gives us a single number representing how "strong" or "long" the vector is.
It's calculated using the Pythagorean theorem in three dimensions. For a vector \( \overrightarrow{A} = a\hat{\imath} + b\hat{\jmath} + c\hat{k} \), the magnitude \( |\overrightarrow{A}| \) can be found using the formula:
  • \( |\overrightarrow{A}| = \sqrt{a^2 + b^2 + c^2} \)

This formula takes into account all three components of the vector: \( a \) in the \( \hat{\imath} \) direction, \( b \) in the \( \hat{\jmath} \) direction, and \( c \) in the \( \hat{k} \) direction. By squaring each component and then taking the square root of their sum, we obtain the vector's magnitude.
For example, for vector \( \overrightarrow{A} = -2.00\hat{\imath} + 3.00\hat{\jmath} + 4.00\hat{k} \), the magnitude would be \( \sqrt{29.00} \approx 5.39 \). This calculation helps in understanding how far the vector reaches in space.
Unit Vectors
Unit vectors are vectors with a magnitude of exactly 1. They're used to indicate direction and are often represented with a 'hat' symbol (\( \hat{\imath}, \hat{\jmath}, \hat{k} \)).
Unit vectors don't change their shape, only pointing out directions in three-dimensional space.
  • The unit vector for \( \overrightarrow{A} = a\hat{\imath} + b\hat{\jmath} + c\hat{k} \) is calculated by dividing each component by the magnitude of the vector:
  • \[ \text{Unit vector of } \overrightarrow{A} = \frac{a}{|\overrightarrow{A}|}\hat{\imath} + \frac{b}{|\overrightarrow{A}|}\hat{\jmath} + \frac{c}{|\overrightarrow{A}|}\hat{k} \]
By using unit vectors, we focus on direction without affecting the vector's length. They often help in simplifying vector calculations, especially in problems involving vector expressions.
Vector Difference
Vector difference is the result of subtracting one vector from another. Subtracting vectors involves subtracting corresponding components.
For instance, to find the difference between vectors \( \overrightarrow{A} \) and \( \overrightarrow{B} \), as shown in the original exercise, we perform the following subtraction:
  • \( \overrightarrow{A} - \overrightarrow{B} = (a_1 - b_1)\hat{\imath} + (a_2 - b_2)\hat{\jmath} + (a_3 - b_3)\hat{k} \)
Here, \( a_1, a_2, a_3 \) and \( b_1, b_2, b_3 \) are the components of \( \overrightarrow{A} \) and \( \overrightarrow{B} \) respectively.
In the example given, the difference \( \overrightarrow{A} - \overrightarrow{B} \) resulted in a new vector \( -5.00\hat{\imath} + 2.00\hat{\jmath} + 7.00\hat{k} \). Calculating the magnitude of this difference provides insights into how "strongly" the direction shifts between the two original vectors.
Vector differences are fundamental in physics and engineering, describing changes or differences in quantities represented by vectors.
Zero Vector
The zero vector is a special vector that has all its components as zero. It is represented as \( \overrightarrow{0} = 0\hat{\imath} + 0\hat{\jmath} + 0\hat{k} \) in three-dimensional space.
Subtraction of any vector from itself results in the zero vector.
  • Example: \( \overrightarrow{A} - \overrightarrow{A} = \overrightarrow{0} \).
  • The magnitude of the zero vector is always zero: \( |\overrightarrow{0}| = 0 \).
Although it doesn't point in any direction, the zero vector serves as an essential concept when dealing with vector arithmetic, ensuring proper mathematical balancing of equations.
Understanding the zero vector is crucial, particularly in linear algebra and vector spaces, as it often acts as a neutral element in vector addition and subtraction.

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