Problem 15 - Calculus Early Transcendentals: Pearson New International Edition

Question

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

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

Step 1: Addition of Vectors

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

Step 2: Subtraction of Vectors

For the difference of the vectors \(\mathbf{u}\) and \(\mathbf{v}\), subtract the corresponding components of \(\mathbf{v}\) from \(\mathbf{u}\): \[\mathbf{u} - \mathbf{v} = \langle 1 - (-5), 0 - 0, 1 - 0 \rangle = \langle 6, 0, 1 \rangle\]

Step 3: Magnitude of Vector u

Calculate the magnitude of \(\mathbf{u}\ = \langle 1, 0, 1 \rangle\) using the formula for the Euclidean norm:\[\|\mathbf{u}\| = \sqrt{1^2 + 0^2 + 1^2} = \sqrt{2}\]

Step 4: Magnitude of Vector v

To find the magnitude of \(\mathbf{v} = \langle -5, 0, 0 \rangle\), apply the same formula:\[\|\mathbf{v}\| = \sqrt{(-5)^2 + 0^2 + 0^2} = \sqrt{25} = 5\]

Key Concepts

Vector Addition

Vector addition is the process of combining two or more vectors to form a new vector. When adding vectors, each component of the vectors involved is added together. For example, given vectors \(\mathbf{u} = \langle 1, 0, 1 \rangle\) and \(\mathbf{v} = \langle -5, 0, 0 \rangle\), their sum can be found by individually adding the components of the vectors.

• Add the x-components: \(1 + (-5) = -4\)
• Add the y-components: \(0 + 0 = 0\)
• Add the z-components: \(1 + 0 = 1\)

Thus, the resulting vector from their addition is \(\langle -4, 0, 1 \rangle\). This new vector represents the combined effect of both vectors \(\mathbf{u}\) and \(\mathbf{v}\) in a three-dimensional space. Vector addition is crucial in physics and engineering where several forces or velocities are acting simultaneously.

Vector Subtraction

Vector subtraction is similar to vector addition but involves taking one vector away from another. For the vectors \(\mathbf{u} = \langle 1, 0, 1 \rangle\) and \(\mathbf{v} = \langle -5, 0, 0 \rangle\), the difference \(\mathbf{u} - \mathbf{v}\) is calculated by subtracting each component of \(\mathbf{v}\) from the respective component of \(\mathbf{u}\).

• Subtract the x-components: \(1 - (-5) = 6\)
• Subtract the y-components: \(0 - 0 = 0\)
• Subtract the z-components: \(1 - 0 = 1\)

This results in the vector \(\langle 6, 0, 1 \rangle\). Vector subtraction is useful in determining the relative difference between two vectors, such as calculating the change in velocity or displacement between two points.

Magnitude of a Vector

The magnitude of a vector represents its length or size and is always a non-negative number. It is calculated using the Euclidean norm formula. For example, to find the magnitude of vector \(\mathbf{u} = \langle 1, 0, 1 \rangle\), you use the following steps.

• Square each component: \(1^2,\ 0^2,\ 1^2\)
• Add these squares: \(1 + 0 + 1 = 2\)
• Take the square root of the sum: \(\sqrt{2}\)

Thus, the magnitude of \(\mathbf{u}\) is \(\sqrt{2}\). Checking the magnitude helps understand the scale or extent of a vector.

Euclidean Norm

The Euclidean norm, often referred to as the magnitude of a vector, is a measure of the vector's length in space. Calculating the Euclidean norm involves finding the positive square root of the sum of the squares of its components. For vector \(\mathbf{v} = \langle -5, 0, 0 \rangle\), we determine the Euclidean norm as follows:

• Square each component: \((-5)^2,\ 0^2,\ 0^2\)
• Sum these squares: \(25 + 0 + 0 = 25\)
• Take the square root of the sum: \(\sqrt{25} = 5\)

The Euclidean norm (magnitude) of \(\mathbf{v}\) is \(5\). It's a crucial concept in vector calculus and is widely used in various applications from computing distances in space to understanding vector relationships.