Problem 16 - 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 \(\|\mathbf{v}\|\). $$ \mathbf{u}=\langle 0.3,0.3,0.5\rangle, \mathbf{v}=\langle 2.2,1.3,-0.9\rangle $$

Step-by-Step Solution

Answer

\( \mathbf{u} + \mathbf{v} = \langle 2.5, 1.6, -0.4 \rangle \), \( \mathbf{u} - \mathbf{v} = \langle -1.9, -1.0, 1.4 \rangle \); \( \| \mathbf{u} \| = 0.655 \), \( \| \mathbf{v} \| = 2.71 \).

Step 1: Finding the Sum of Vectors

To find the sum of vectors \( \mathbf{u} \) and \( \mathbf{v} \), add corresponding components: \( \mathbf{u} + \mathbf{v} = \langle 0.3 + 2.2, 0.3 + 1.3, 0.5 + (-0.9) \rangle \). \[ \mathbf{u} + \mathbf{v} = \langle 2.5, 1.6, -0.4 \rangle \]

Step 2: Finding the Difference of Vectors

To find the difference \( \mathbf{u} - \mathbf{v} \), subtract corresponding components of \( \mathbf{v} \) from \( \mathbf{u} \):\[ \mathbf{u} - \mathbf{v} = \langle 0.3 - 2.2, 0.3 - 1.3, 0.5 - (-0.9) \rangle \] \[ \mathbf{u} - \mathbf{v} = \langle -1.9, -1.0, 1.4 \rangle \]

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

To find the magnitude of \( \mathbf{u} \), use the formula:\[ \| \mathbf{u} \| = \sqrt{(0.3)^2 + (0.3)^2 + (0.5)^2} \]\[ \| \mathbf{u} \| = \sqrt{0.09 + 0.09 + 0.25} = \sqrt{0.43} = 0.655 \]

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

To find the magnitude of \( \mathbf{v} \), use the formula:\[ \| \mathbf{v} \| = \sqrt{(2.2)^2 + (1.3)^2 + (-0.9)^2} \]\[ \| \mathbf{v} \| = \sqrt{4.84 + 1.69 + 0.81} = \sqrt{7.34} = 2.71 \]

Step 5: Conclusion: Final Results Compilation

The sum of \( \mathbf{u} \) and \( \mathbf{v} \) is \( \langle 2.5, 1.6, -0.4 \rangle \); the difference is \( \langle -1.9, -1.0, 1.4 \rangle \). The magnitudes are \( \| \mathbf{u} \| = 0.655 \) and \( \| \mathbf{v} \| = 2.71 \).

Key Concepts

Magnitude of a Vector

Vectors have a length, called their magnitude, which is a scalar quantity. It's essentially a measure of how long or large the vector is in a geometric sense.
It's calculated using the Pythagorean theorem adapted to three dimensions.
The magnitude of a vector \( \mathbf{a} = \langle x, y, z \rangle \) is determined using:

• \( \| \mathbf{a} \| = \sqrt{x^2 + y^2 + z^2} \)

For example, for the vector \( \mathbf{u} = \langle 0.3, 0.3, 0.5 \rangle \), we find its magnitude by calculating:
\[ \| \mathbf{u} \| = \sqrt{(0.3)^2 + (0.3)^2 + (0.5)^2} = \sqrt{0.09 + 0.09 + 0.25} = \sqrt{0.43} \approx 0.655 \]
This value represents how far the point is from the origin in a three-dimensional space.

Three-Dimensional Vectors

Three-dimensional vectors are used to describe quantities having both a magnitude and direction in three-dimensional space. Each vector can be represented by its components with respect to the \( x \), \( y \), and \( z \) axes.
For instance, the vector \( \mathbf{u} = \langle 0.3, 0.3, 0.5 \rangle \) means it extends 0.3 units along the x-axis, 0.3 units along the y-axis, and 0.5 units along the z-axis. This representation enables us to understand the vector’s precise position and direction.

• Such vectors are fundamental in physics and engineering, representing forces, velocities, and other vector quantities in 3D space.
• The ability to express vectors in three dimensions allows us to perform operations like addition, subtraction, and more, maintaining each component's coordinates.

This facilitates complex geometrical and physical computations with simple algebraic operations.

Vector Subtraction

Subtracting vectors involves taking one vector away from another, which entails subtracting the individual corresponding components.
When given two vectors, \( \mathbf{u} = \langle 0.3, 0.3, 0.5 \rangle \) and \( \mathbf{v} = \langle 2.2, 1.3, -0.9 \rangle \), the subtraction \( \mathbf{u} - \mathbf{v} \) adjusts each component separately:

• Subtract their \( x \)-components: \( 0.3 - 2.2 = -1.9 \)
• Subtract their \( y \)-components: \( 0.3 - 1.3 = -1.0 \)
• Subtract their \( z \)-components: \( 0.5 - (-0.9) = 1.4 \)

Thus, the result is \( \mathbf{u} - \mathbf{v} = \langle -1.9, -1.0, 1.4 \rangle \).
This operation is similar to vector addition and follows the principle of superposition, which is essential in analyzing forces and other vector quantities.