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Chapter 1: Problem 27

For the following exercises, vectors \(\mathbf{u}\) and \(\mathbf{v}\) are given. Find the magnitudes of vectors \(\mathbf{u}-\mathbf{v}\) and \(-2 \mathbf{u}\). $$ \mathbf{u}=2 \mathbf{i}+3 \mathbf{j}+4 \mathbf{k}, \mathbf{v}=-\mathbf{i}+5 \mathbf{j}-\mathbf{k} $$

Short Answer

Expert verified
The magnitudes are \( \sqrt{38} \) for \( \mathbf{u} - \mathbf{v} \) and \( \sqrt{116} \) for \(-2\mathbf{u} \).

Step by step solution

01

Calculate \( \mathbf{u} - \mathbf{v} \)

To find \( \mathbf{u} - \mathbf{v} \), subtract the corresponding components of \( \mathbf{u} \) and \( \mathbf{v} \). Thus, \( \mathbf{u} - \mathbf{v} = (2 - (-1))\mathbf{i} + (3 - 5)\mathbf{j} + (4 - (-1))\mathbf{k} \). Simplifying gives \( \mathbf{u} - \mathbf{v} = 3\mathbf{i} - 2\mathbf{j} + 5\mathbf{k} \).
02

Find the magnitude of \( \mathbf{u} - \mathbf{v} \)

The magnitude of a vector \( \mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k} \) is given by \( \sqrt{a_1^2 + a_2^2 + a_3^2} \). For \( 3\mathbf{i} - 2\mathbf{j} + 5\mathbf{k} \), the magnitude is \( \sqrt{3^2 + (-2)^2 + 5^2} = \sqrt{9 + 4 + 25} = \sqrt{38} \).
03

Calculate \( -2 \mathbf{u} \)

To find \( -2 \mathbf{u} \), multiply each component of \( \mathbf{u} = 2 \mathbf{i} + 3 \mathbf{j} + 4 \mathbf{k} \) by -2. This results in \( -2 \mathbf{u} = -4 \mathbf{i} - 6 \mathbf{j} - 8 \mathbf{k} \).
04

Find the magnitude of \( -2 \mathbf{u} \)

The magnitude of \( -2 \mathbf{u} = -4 \mathbf{i} - 6 \mathbf{j} - 8 \mathbf{k} \) is calculated as \( \sqrt{(-4)^2 + (-6)^2 + (-8)^2} = \sqrt{16 + 36 + 64} = \sqrt{116} \).

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Key Concepts

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

Vector Subtraction
When working in vector calculus, vector subtraction is an essential concept. It allows us to find the difference between two vectors by subtracting their corresponding components. First, let's consider two vectors, \( \mathbf{u} = 2 \mathbf{i} + 3 \mathbf{j} + 4 \mathbf{k} \) and \( \mathbf{v} = -\mathbf{i} + 5 \mathbf{j} - \mathbf{k} \). To subtract \( \mathbf{v} \) from \( \mathbf{u} \), we compute:
  • Subtract the \( \mathbf{i} \) components: \( 2 - (-1) = 3 \)
  • Subtract the \( \mathbf{j} \) components: \( 3 - 5 = -2 \)
  • Subtract the \( \mathbf{k} \) components: \( 4 - (-1) = 5 \)
Thus, \( \mathbf{u} - \mathbf{v} = 3\mathbf{i} - 2\mathbf{j} + 5\mathbf{k} \).This result shows that vector subtraction is simply performing arithmetic operations on each dimension separately, producing a new vector that reflects how \( \mathbf{u} \) and \( \mathbf{v} \) differ.
Vector Magnitude
The magnitude of a vector represents its "length" in space and is an important aspect in vector calculus. The formula for the magnitude of a vector \( \mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k} \) is \( \sqrt{a_1^2 + a_2^2 + a_3^2} \).To find the magnitude of our resultant vector \( \mathbf{u} - \mathbf{v} = 3\mathbf{i} - 2\mathbf{j} + 5\mathbf{k} \):
  • Square each component: \( 3^2 = 9 \), \((-2)^2 = 4 \), \(5^2 = 25\)
  • Add the squares: \( 9 + 4 + 25 = 38 \)
  • Take the square root: \( \sqrt{38} \)
The magnitude \( \sqrt{38} \) informs us about the size of the vector in 3D space. The process of finding a vector's magnitude essentially determines how long or "intense" the vector is.
Scalar Multiplication
Scalar multiplication is a basic operation in vector calculus, allowing us to scale a vector by a fixed number, known as a scalar. When a vector is multiplied by a scalar, each of its components is multiplied by that scalar. Consider the vector \( \mathbf{u} = 2\mathbf{i} + 3\mathbf{j} + 4\mathbf{k} \) and the scalar \(-2\). To find \(-2 \mathbf{u} \), perform the following:
  • Multiply the \( \mathbf{i} \) component: \( -2 \times 2 = -4 \)
  • Multiply the \( \mathbf{j} \) component: \( -2 \times 3 = -6 \)
  • Multiply the \( \mathbf{k} \) component: \( -2 \times 4 = -8 \)
The result is \(-2 \mathbf{u} = -4\mathbf{i} - 6\mathbf{j} - 8\mathbf{k} \). This operation affects both the direction and the size of the vector. When the scalar is negative, the vector's direction is reversed.

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