Problem 13 - 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 \(\|\mathbf{v}\|\). $$ \mathbf{u}=\langle-1,0,0\rangle, \mathbf{v}=\langle 3,4,0\rangle $$

Step-by-Step Solution

Answer

The sum is \( \langle 2, 4, 0 \rangle \), the difference is \( \langle -4, -4, 0 \rangle \), and the magnitudes are 1 and 5.

Step 1: Sum of Vectors

To find the sum of vectors \( \mathbf{u} \) and \( \mathbf{v} \), add their corresponding components. \(\mathbf{u} + \mathbf{v} = \langle -1+3, 0+4, 0+0 \rangle = \langle 2, 4, 0 \rangle\).

Step 2: Difference of Vectors

To find the difference \( \mathbf{u} - \mathbf{v} \), subtract the components of \( \mathbf{v} \) from \( \mathbf{u} \). \(\mathbf{u} - \mathbf{v} = \langle -1-3, 0-4, 0-0 \rangle = \langle -4, -4, 0 \rangle\).

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

The magnitude of vector \( \mathbf{u} = \langle -1, 0, 0 \rangle \) is calculated using the formula \( \|\mathbf{u}\| = \sqrt{(-1)^2 + 0^2 + 0^2} = \sqrt{1} = 1 \).

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

The magnitude of vector \( \mathbf{v} = \langle 3, 4, 0 \rangle \) is calculated using the formula \( \|\mathbf{v}\| = \sqrt{3^2 + 4^2 + 0^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \).

Key Concepts

Sum of Vectors

Vector addition is quite simple. It involves combining two or more vectors to get a resultant vector, also known as the sum. To find the sum of two vectors, you add their respective components together.
For instance, when you have the vectors \( \mathbf{u} = \langle -1, 0, 0 \rangle \) and \( \mathbf{v} = \langle 3, 4, 0 \rangle \), you add each component of \( \mathbf{u} \) to the corresponding component of \( \mathbf{v} \).

This is done as follows:

• For the first component: \(-1 + 3 = 2 \)
• For the second component: \(0 + 4 = 4 \)
• For the third component: \(0 + 0 = 0 \)

Thus, the sum \( \mathbf{u} + \mathbf{v} \) becomes \( \langle 2, 4, 0 \rangle \).
Each component is treated independently, making vector addition straightforward and easy to apply across different contexts.

Difference of Vectors

Vector subtraction, like vector addition, involves component-wise operations. To find the difference between two vectors, we subtract the components of the second vector from the first.
This subtraction is done as follows for vectors \( \mathbf{u} \) and \( \mathbf{v} \):

• For the first component: \(-1 - 3 = -4 \)
• For the second component: \(0 - 4 = -4 \)
• For the third component: \(0 - 0 = 0 \)

Consequently, the difference \( \mathbf{u} - \mathbf{v} \) results in \( \langle -4, -4, 0 \rangle \).

This subtraction effectively reflects how one vector differs directionally and in magnitude from another. The operation is basic yet crucial for understanding the spatial relationship between vectors.

Magnitude of Vectors

The magnitude of a vector is like measuring its length in space. It's how we determine the size of the vector, without regard for direction. This is computed using the Pythagorean theorem.
For vectors in three dimensions, such as \( \mathbf{u} = \langle -1, 0, 0 \rangle \) and \( \mathbf{v} = \langle 3, 4, 0 \rangle \), the magnitude is calculated as follows:

• For \( \mathbf{u}: \|\mathbf{u} \| = \sqrt{(-1)^2 + 0^2 + 0^2} = \sqrt{1} = 1 \)
• For \( \mathbf{v}: \|\mathbf{v} \| = \sqrt{3^2 + 4^2 + 0^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \)

These steps show that the magnitude provides a scalar depicting the distance from the origin to the point described by the vector.
It is essential as it gives insights into the vector's "strength" or "intensity" and is widely used in physics and engineering to understand vector behavior.