Problem 41 - Precalculus with Calculus Previews

Question

Normalize the given vector when \(\mathbf{v}=\langle 2,8\rangle\) and \(\mathbf{w}=\langle 3,4\rangle\). $$ \mathbf{V}+\mathbf{W} $$

Step-by-Step Solution

Answer

The normalized vector is \( \langle \frac{5}{13}, \frac{12}{13} \rangle \)."

Step 1: Find Vector Sum

Add the vectors \( \mathbf{v} = \langle 2,8 \rangle \) and \( \mathbf{w} = \langle 3,4 \rangle \). This is done component-wise: \( \mathbf{v} + \mathbf{w} = \langle 2+3, 8+4 \rangle = \langle 5, 12 \rangle \).

Step 2: Compute Magnitude

Compute the magnitude of the resulting vector \( \langle 5, 12 \rangle \). The magnitude is found using the formula \( \| \langle 5, 12 \rangle \| = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \).

Step 3: Normalize the Vector

To normalize the vector, divide each component by its magnitude. So the normalized vector is \( \langle \frac{5}{13}, \frac{12}{13} \rangle \).

Key Concepts

Vector Addition

Vector addition is a method used to combine two vectors, resulting in a new vector. This process is usually performed using the component-wise approach, which means that each corresponding component of the two vectors is added together. For the vectors \( \mathbf{v} = \langle 2, 8 \rangle \) and \( \mathbf{w} = \langle 3, 4 \rangle \), the addition process looks like this:
\( \mathbf{v} + \mathbf{w} = \langle 2+3, 8+4 \rangle = \langle 5, 12 \rangle \).

• The first component of \( \mathbf{v} \) and \( \mathbf{w} \) are 2 and 3. Adding them gives 5.
• The second components, 8 and 4, add up to 12.

This component-wise technique allows us to handle vectors as simple numbers, making calculations straightforward and intuitive. The resulting vector \( \langle 5, 12 \rangle \) is the sum of the original two vectors.

Magnitude of a Vector

The magnitude of a vector, also known as its length or norm, is a measure of how long the vector is. It's calculated using the Pythagorean Theorem, which is applied to the vector's components. If a vector is expressed as \( \langle x, y \rangle \), its magnitude \( \| \langle x, y \rangle \| \) is found using the square root of the sum of the squares of its components:
\[ \| \langle x, y \rangle \| = \sqrt{x^2 + y^2} \]
Let's apply this to our vector \( \langle 5, 12 \rangle \):

• Square each component: 5 becomes 25, and 12 becomes 144.
• Add these squares: 25 + 144 = 169.
• Calculate the square root of this sum to get the magnitude: \( \sqrt{169} = 13 \).

The magnitude tells us the total "distance" covered by the vector, starting from the origin \( \langle 0, 0 \rangle \) to the point \( \langle x, y \rangle \). In this example, the vector \( \langle 5, 12 \rangle \) has a magnitude of 13, indicating the vector stretches across this length in the two-dimensional plane.

Component-wise Operations

Component-wise operations are fundamental to vector arithmetic and involve working with each part or 'component' of the vectors separately. This technique simplifies various operations like addition, subtraction, and even normalization, as each component is treated as an independent entity.

For vector normalization, which means resizing the vector to have a magnitude of 1, you perform component-wise division. This is done by dividing each component by the vector's magnitude. For the vector \( \langle 5, 12 \rangle \) with a magnitude of 13, normalization is achieved by calculating:
\[ \left( \frac{5}{13}, \frac{12}{13} \right) \]
During this operation:

• Each component of the vector, 5 and 12, is divided by 13.
• This results in the normalized vector components \( \frac{5}{13} \) and \( \frac{12}{13} \).

Component-wise operations allow vectors to be manipulated in a straightforward manner. This is particularly useful when dealing with real-world physics and engineering problems where exact vector directions are crucial.