Problem 14 - 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 \(\|\mathbf{v}\|\) $$ \mathbf{u}=\langle 0,0,0\rangle, \mathbf{v}=\langle-3,3,1\rangle $$

Step-by-Step Solution

Answer

\( \mathbf{u} + \mathbf{v} = \langle -3, 3, 1 \rangle \), \( \mathbf{u} - \mathbf{v} = \langle 3, -3, -1 \rangle \); \( \|\mathbf{u}\| = 0 \), \( \|\mathbf{v}\| = \sqrt{19} \).

Step 1: Sum of Vectors

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

Step 2: Difference of Vectors

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

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

The magnitude of a vector \( \langle a, b, c \rangle \) is given by \( \|\mathbf{u}\| = \sqrt{a^2 + b^2 + c^2} \). For \( \mathbf{u} = \langle 0, 0, 0 \rangle \), we calculate \( \|\mathbf{u}\| = \sqrt{0^2 + 0^2 + 0^2} = 0 \).

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

Similarly, calculate the magnitude of \( \mathbf{v} = \langle -3, 3, 1 \rangle \). Using the formula, \( \|\mathbf{v}\| = \sqrt{(-3)^2 + 3^2 + 1^2} = \sqrt{9 + 9 + 1} = \sqrt{19} \).

Key Concepts

Understanding Vector Addition

Vector addition is a basic operation in vector mathematics that involves combining two vectors to form a new vector. To add two vectors, you simply add their corresponding components together. This process is quite straightforward:

• Consider vectors \( \mathbf{u} = \langle u_1, u_2, u_3 \rangle \) and \( \mathbf{v} = \langle v_1, v_2, v_3 \rangle \).
• The sum \( \mathbf{u} + \mathbf{v} \) results in a new vector \( \langle u_1 + v_1, u_2 + v_2, u_3 + v_3 \rangle \).

In our example, our vectors are \( \mathbf{u} = \langle 0, 0, 0 \rangle \) and \( \mathbf{v} = \langle -3, 3, 1 \rangle \). Adding these, we calculate:\[ \mathbf{u} + \mathbf{v} = \langle 0 + (-3), 0 + 3, 0 + 1 \rangle = \langle -3, 3, 1 \rangle \]This result is quite manageable: simply take each component from both vectors and add them together.

The result, \( \langle -3, 3, 1 \rangle \), is the new vector which shows the combined effect of the two original vectors.

Understanding Vector Subtraction

Vector subtraction is similar to vector addition, but instead of adding components, you subtract them. This operation finds the vector that points from one vector to another. Let's break it down:

• For vectors \( \mathbf{u} = \langle u_1, u_2, u_3 \rangle \) and \( \mathbf{v} = \langle v_1, v_2, v_3 \rangle \), the difference \( \mathbf{u} - \mathbf{v} \) is \( \langle u_1 - v_1, u_2 - v_2, u_3 - v_3 \rangle \).

Examining our example, where \( \mathbf{u} = \langle 0, 0, 0 \rangle \) and \( \mathbf{v} = \langle -3, 3, 1 \rangle \), the subtraction is computed as:\[\mathbf{u} - \mathbf{v} = \langle 0 - (-3), 0 - 3, 0 - 1 \rangle = \langle 3, -3, -1 \rangle\]

By performing vector subtraction, you can determine how the vector \( \mathbf{u} \) changes in relation to vector \( \mathbf{v} \). The result \( \langle 3, -3, -1 \rangle \) denotes this relative change, illustrating the direction and magnitude of difference between the two vectors.

Determining the Magnitude of a Vector

The magnitude of a vector (also known as its length) is a measure of its size in space. For any vector \( \mathbf{v} = \langle a, b, c \rangle \), its magnitude is given by the formula:\[\| \mathbf{v} \| = \sqrt{a^2 + b^2 + c^2}\]This formula resembles the Pythagorean theorem, which you might recognize from geometry.

• It calculates the hypotenuse of a right triangle formed by the vector components.

For vector \( \mathbf{u} = \langle 0, 0, 0 \rangle \), applying the formula:\[\| \mathbf{u} \| = \sqrt{0^2 + 0^2 + 0^2} = 0\]This means vector \( \mathbf{u} \) has no length, which is logical for a zero vector.

Now, consider vector \( \mathbf{v} = \langle -3, 3, 1 \rangle \). Its magnitude is calculated as:\[\| \mathbf{v} \| = \sqrt{(-3)^2 + 3^2 + 1^2} = \sqrt{9 + 9 + 1} = \sqrt{19}\]
Therefore, the length of \( \mathbf{v} \) in space is approximately 4.36. Understanding vector magnitude is crucial for describing how vectors behave, representing their strength and direction effectively.