Problem 15 - Calculus Early Transcendentals: Pearson New International Edition

Question

For the three-dimensional vectors \(\mathbf{u}\) and \(\mathbf{v}\) in Problems 13-16, find the sum \(\mathbf{u}+\mathbf{v}\), the difference \(\mathbf{u}-\mathbf{v}\), and the magnitudes \(\|\mathbf{u}\|\) and \(\|\mathbf{v}\|\). $$ \mathbf{u}=\langle 1,0,1\rangle, \mathbf{v}=\langle-5,0,0\rangle $$

Step-by-Step Solution

Answer

\(\mathbf{u} + \mathbf{v} = \langle -4, 0, 1 \rangle\), \(\mathbf{u} - \mathbf{v} = \langle 6, 0, 1 \rangle\), \(\|\mathbf{u}\| = \sqrt{2}\), \(\|\mathbf{v}\| = 5\).

Step 1: Calculate the Sum of the Vectors

To find the sum of the vectors \(\mathbf{u}\) and \(\mathbf{v}\), add their corresponding components. \[ \mathbf{u} + \mathbf{v} = \langle 1, 0, 1 \rangle + \langle -5, 0, 0 \rangle = \langle 1 + (-5), 0 + 0, 1 + 0 \rangle = \langle -4, 0, 1 \rangle \]

Step 2: Calculate the Difference of the Vectors

To find the difference between the vectors \(\mathbf{u}\) and \(\mathbf{v}\), subtract the components of \(\mathbf{v}\) from \(\mathbf{u}\).\[ \mathbf{u} - \mathbf{v} = \langle 1, 0, 1 \rangle - \langle -5, 0, 0 \rangle = \langle 1 - (-5), 0 - 0, 1 - 0 \rangle = \langle 6, 0, 1 \rangle \]

Step 3: Calculate the Magnitude of Vector \(\mathbf{u}\)

The magnitude of a vector \(\mathbf{u} = \langle x, y, z \rangle\) is given by \[ \|\mathbf{u}\| = \sqrt{x^2 + y^2 + z^2} \] Substitute the components of \(\mathbf{u}\):\[ \|\mathbf{u}\| = \sqrt{1^2 + 0^2 + 1^2} = \sqrt{1 + 0 + 1} = \sqrt{2} \]

Step 4: Calculate the Magnitude of Vector \(\mathbf{v}\)

The magnitude of a vector \(\mathbf{v} = \langle x, y, z \rangle\) is calculated the same way:\[ \|\mathbf{v}\| = \sqrt{x^2 + y^2 + z^2} \] Substitute the components of \(\mathbf{v}\):\[ \|\mathbf{v}\| = \sqrt{(-5)^2 + 0^2 + 0^2} = \sqrt{25 + 0 + 0} = \sqrt{25} = 5 \]

Key Concepts

Vector Addition

Vector addition is a fundamental concept in vector calculus that involves combining two or more vectors to produce a resultant vector. To add vectors, simply add their corresponding components.Imagine each vector as an arrow pointing in space from a starting point to an endpoint. When you add vectors together, you're essentially joining these arrows end-to-end.Consider vectors \( \mathbf{u} = \langle 1, 0, 1 \rangle \) and \( \mathbf{v} = \langle -5, 0, 0 \rangle \). To find \( \mathbf{u} + \mathbf{v} \), follow these steps:

• Add the first components: \( 1 + (-5) = -4 \)
• Add the second components: \( 0 + 0 = 0 \)
• Add the third components: \( 1 + 0 = 1 \)

Thus, the resultant vector is \( \langle -4, 0, 1 \rangle \). This vector represents the overall effect of applying both \( \mathbf{u} \) and \( \mathbf{v} \).

Vector Subtraction

Vector subtraction is the process of finding the change in one vector relative to another. This is done by subtracting the corresponding components of the vectors.Imagine vector subtraction as determining the difference in direction and magnitude between two arrows representing the vectors.For vectors \( \mathbf{u} = \langle 1, 0, 1 \rangle \) and \( \mathbf{v} = \langle -5, 0, 0 \rangle \), the difference \( \mathbf{u} - \mathbf{v} \) can be calculated by:

• Subtracting the first components: \( 1 - (-5) = 6 \)
• Subtracting the second components: \( 0 - 0 = 0 \)
• Subtracting the third components: \( 1 - 0 = 1 \)

The resulting vector is \( \langle 6, 0, 1 \rangle \). This represents how \( \mathbf{u} \) differs from \( \mathbf{v} \) in both direction and length.

Magnitude of a Vector

The magnitude of a vector, often represented as \( \| \mathbf{a} \| \), is a measure of its length. It's calculated using the Pythagorean theorem for three-dimensional space.The formula to find the magnitude of a vector \( \mathbf{a} = \langle x, y, z \rangle \) is: \[ \| \mathbf{a} \| = \sqrt{x^2 + y^2 + z^2} \]For vector \( \mathbf{u} = \langle 1, 0, 1 \rangle \), the magnitude is:

• \( \| \mathbf{u} \| = \sqrt{1^2 + 0^2 + 1^2} = \sqrt{1 + 0 + 1} = \sqrt{2} \)

Similarly, for vector \( \mathbf{v} = \langle -5, 0, 0 \rangle \), calculate:

• \( \| \mathbf{v} \| = \sqrt{(-5)^2 + 0^2 + 0^2} = \sqrt{25} = 5 \)

These magnitudes represent the lengths or sizes of the vectors in three-dimensional space.

Three-Dimensional Vectors

Three-dimensional vectors are vectors that have components in three directions, typically denoted as \( x, y, \) and \( z \). These components help describe an object's position or movement in space with respect to a reference point.Vectors \( \mathbf{u} = \langle 1, 0, 1 \rangle \) and \( \mathbf{v} = \langle -5, 0, 0 \rangle \) each have three components:

• The \( x \)-component shows horizontal movement.
• The \( y \)-component shows vertical movement.
• The \( z \)-component typically shows depth or movement in/out of a plane.

Understanding three-dimensional vectors is crucial in physics and engineering, as they allow us to describe forces, velocities, and other quantities directionally in space. These concepts help us visualize and solve real-world problems involving spatial relationships. By analyzing each component, we can determine how objects interact within this three-dimensional space.