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Given two vectors \(\overrightarrow{\boldsymbol{A}}=-2.00 \hat{\mathfrak{i}}+3.00 \hat{\mathbf{j}}+4.00 \hat{\boldsymbol{k}}\) and \(\overrightarrow{\boldsymbol{B}}=\) \(3.00 \hat{t}+1.00 \hat{\jmath}-3.00 \hat{k},\) do the following. (a) Find the magnitude of each vector. (b) Write an expression for the vector difference \(\vec{A}-\vec{B},\) using unit vectors. (c) Find the magnitude of the vector difference \(\overrightarrow{\boldsymbol{A}}-\overrightarrow{\boldsymbol{B}} .\) Is this the same as the magnitude of \(\overrightarrow{\boldsymbol{B}}-\overrightarrow{\boldsymbol{A}} ?\) Explain.

Short Answer

Expert verified
Magnitudes: |A| \(\approx 5.39\), |B| \(\approx 4.36\). A-B: \(-5\hat{i} + 2\hat{j} + 7\hat{k}\), |A-B| \(\approx 8.83\). Magnitudes of A-B and B-A are equal.

Step by step solution

01

Calculate the Magnitude of Vector A

The magnitude of a vector \( \overrightarrow{\boldsymbol{A}} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k} \) is calculated as \( \sqrt{a_1^2 + a_2^2 + a_3^2} \). For vector \( \overrightarrow{\boldsymbol{A}} = -2.00 \hat{i} + 3.00 \hat{j} + 4.00 \hat{k} \), the magnitude is:\[ |\overrightarrow{\boldsymbol{A}}| = \sqrt{(-2.00)^2 + (3.00)^2 + (4.00)^2} = \sqrt{4.00 + 9.00 + 16.00} = \sqrt{29.00} \approx 5.39 \]
02

Calculate the Magnitude of Vector B

Similarly, for vector \( \overrightarrow{\boldsymbol{B}} = 3.00 \hat{i} + 1.00 \hat{j} - 3.00 \hat{k} \), the magnitude is:\[ |\overrightarrow{\boldsymbol{B}}| = \sqrt{(3.00)^2 + (1.00)^2 + (-3.00)^2} = \sqrt{9.00 + 1.00 + 9.00} = \sqrt{19.00} \approx 4.36 \]
03

Calculate the Vector Difference A - B

The vector difference \( \overrightarrow{\boldsymbol{A}} - \overrightarrow{\boldsymbol{B}} \) is given by subtracting the corresponding components of \( \overrightarrow{\boldsymbol{A}} \) and \( \overrightarrow{\boldsymbol{B}} \):\[ \overrightarrow{\boldsymbol{A}} - \overrightarrow{\boldsymbol{B}} = (-2.00 - 3.00) \hat{i} + (3.00 - 1.00) \hat{j} + (4.00 - (-3.00)) \hat{k} \] This simplifies to:\[ \overrightarrow{\boldsymbol{A}} - \overrightarrow{\boldsymbol{B}} = -5.00 \hat{i} + 2.00 \hat{j} + 7.00 \hat{k} \]
04

Calculate the Magnitude of Vector Difference A - B

The magnitude of the vector difference \( \overrightarrow{\boldsymbol{A}} - \overrightarrow{\boldsymbol{B}} = -5.00 \hat{i} + 2.00 \hat{j} + 7.00 \hat{k} \) is:\[ |\overrightarrow{\boldsymbol{A}} - \overrightarrow{\boldsymbol{B}}| = \sqrt{(-5.00)^2 + (2.00)^2 + (7.00)^2} = \sqrt{25.00 + 4.00 + 49.00} = \sqrt{78.00} \approx 8.83 \]
05

Discuss Magnitude A - B vs B - A

The magnitude of \( \overrightarrow{\boldsymbol{B}} - \overrightarrow{\boldsymbol{A}} \) is the same as \( \overrightarrow{\boldsymbol{A}} - \overrightarrow{\boldsymbol{B}} \) because magnitude is a scalar and does not depend on direction:\[ |\overrightarrow{\boldsymbol{B}} - \overrightarrow{\boldsymbol{A}}| = |\overrightarrow{\boldsymbol{A}} - \overrightarrow{\boldsymbol{B}}| = \sqrt{(-5.00)^2 + (-2.00)^2 + (-7.00)^2} = \sqrt{78.00} \approx 8.83 \]Thus, the magnitude is the same for both vector differences.

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Key Concepts

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

Vector Operations
Vector operations are fundamental in physics and mathematics, helping us manipulate vectors in a wide range of applications. Vectors are entities having both magnitude and direction, typically used to represent quantities like force, velocity, and displacement.

When performing vector operations, such as addition, subtraction, or scalar multiplication, we must consider each component of the vectors involved. This means working with the individual parts that make up the vector in the coordinate system, often denoted as components along the x, y, and z axes.

To illustrate vector subtraction, consider two vectors, \( \overrightarrow{A} \) and \( \overrightarrow{B} \). To subtract \( \overrightarrow{B} \) from \( \overrightarrow{A} \), denoted as \( \overrightarrow{A} - \overrightarrow{B} \), you subtract the corresponding components from each other:
  • \( x \)-component: \( a_1 - b_1 \)
  • \( y \)-component: \( a_2 - b_2 \)
  • \( z \)-component: \( a_3 - b_3 \)
This process allows us to find a new vector that represents the direction and magnitude of the difference between the two original vectors.
Magnitude Calculation
Magnitude calculation helps us determine the size or length of a vector. This is crucial when we talk about the strength or distance related to a vector quantity.

The magnitude of a vector \( \overrightarrow{A} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k} \) can be calculated using the Pythagorean theorem for three dimensions, given by the formula:
\[ |\overrightarrow{A}| = \sqrt{a_1^2 + a_2^2 + a_3^2} \]
This calculation gives us a scalar, a quantity without direction, which tells us how long the vector is.

For example, consider a vector \( \overrightarrow{A} = -2.00 \hat{i} + 3.00 \hat{j} + 4.00 \hat{k} \). The magnitude can be computed as \( \sqrt{(-2.00)^2 + (3.00)^2 + (4.00)^2} = \sqrt{29.00} \approx 5.39 \).

This indicates the length of the vector in space, irrespective of its direction. Calculating magnitudes is essential in physics to understand quantities like acceleration or forces, focusing more on their size rather than their orientation.
Unit Vectors
Unit vectors play a significant role in describing directions. A unit vector has a magnitude of 1 and indicates direction along one of the axes in a coordinate system.

When we want to express a vector in terms of direction, we often normalize it. Normalization is the process of converting the vector into a unit vector by dividing each component of the vector by its magnitude.

For example, if you have a vector \( \overrightarrow{C} = (a, b, c) \), its unit vector \( \hat{C} \) can be derived as:
\[ \hat{C} = \frac{1}{|\overrightarrow{C}|} \overrightarrow{C} = \left( \frac{a}{|\overrightarrow{C}|}, \frac{b}{|\overrightarrow{C}|}, \frac{c}{|\overrightarrow{C}|} \right) \]

Unit vectors are often used in combination with other operations in vector mathematics to define directions along the x, y, and z axes using standard notations \( \hat{i} \), \( \hat{j} \), and \( \hat{k} \). These are helpful in performing projections of vectors on given lines or planes or for simplifying complex vector calculations by focusing solely on direction.

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Most popular questions from this chapter

Bond Angle in Methane. In the methane molecule, \(\mathbf{C H}_{4}\), each hydrogen atom is at a corner of a regular terrahedron with the carbon atom at the center. In coordinates where one of the \(C-H\) bonds is in the direction of \(\hat{\imath}+\hat{\jmath}+\hat{k},\) an adjacent \(\mathbf{C}-\mathbf{H}\) bond is in the \(\hat{\imath}-\hat{\jmath}-\hat{k}\) direction. Calculate the angle between these two bonds.

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A disoriented physics professor drives 3.25 \(\mathrm{km}\) north, then 4.75 \(\mathrm{km}\) west, and then 1.50 \(\mathrm{km}\) south. Find the magnitude and direction of the resultant displacement, using the method of components. In a vector addition diagram (roughly to scale), show that the resultant displacement found from your diagram is in qualitative agreement with the result you obtained using the method of components.

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