Problem 42 - Precalculus with Calculus Previews

Question

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

Step-by-Step Solution

Answer

The normalized vector is \( \left\langle \frac{-5}{\sqrt{41}}, \frac{4}{\sqrt{41}} \right\rangle \).

Step 1: Evaluate the Expression

The given expression is a combination of vectors \( \mathbf{v} = \langle 2, 8 \rangle \) and \( \mathbf{w} = \langle 3, 4 \rangle \). The expression is \( 2\mathbf{v} - 3\mathbf{w} \). First, we calculate \( 2\mathbf{v} \) and \( 3\mathbf{w} \).\[ 2\mathbf{v} = 2 \times \langle 2, 8 \rangle = \langle 4, 16 \rangle \] \[ 3\mathbf{w} = 3 \times \langle 3, 4 \rangle = \langle 9, 12 \rangle \].Next, substitute these into the expression to get: \[ 2\mathbf{v} - 3\mathbf{w} = \langle 4, 16 \rangle - \langle 9, 12 \rangle = \langle 4-9, 16-12 \rangle = \langle -5, 4 \rangle \].

Step 2: Calculate the Magnitude of the Resultant Vector

The resultant vector from Step 1 is \( \langle -5, 4 \rangle \). To normalize it, we first need to find its magnitude. The formula for the magnitude of a vector \( \langle a, b \rangle \) is \( \sqrt{a^2 + b^2} \). Substitute the values:\[ \text{Magnitude} = \sqrt{(-5)^2 + 4^2} = \sqrt{25 + 16} = \sqrt{41} \].

Step 3: Normalize the Vector

To normalize the vector \( \langle -5, 4 \rangle \), divide each component by the magnitude calculated in Step 2. The normalized vector \( \mathbf{u} \) is given by:\[ \mathbf{u} = \frac{1}{\sqrt{41}} \cdot \langle -5, 4 \rangle = \left\langle \frac{-5}{\sqrt{41}}, \frac{4}{\sqrt{41}} \right\rangle \].

Key Concepts

Vector Operations

Vectors are essential tools in mathematics and physics. They represent quantities that have both magnitude and direction. Understanding vector operations helps in various applications, from physics to engineering. Common operations include addition, subtraction, and scalar multiplication. For instance, when we have vectors like \( \mathbf{v} = \langle 2, 8 \rangle \) and \( \mathbf{w} = \langle 3, 4 \rangle \), we can perform operations such as scalar multiplication and addition or subtraction of vectors.

Scalar multiplication involves multiplying each component of a vector by a scalar (a real number). For example, multiplying vector \( \mathbf{v} \) by 2 yields \( 2\mathbf{v} = \langle 4, 16 \rangle \).

• **Addition:** Combine corresponding components from two vectors.
• **Subtraction:** Subtract corresponding components of one vector from another.
• **Scalar multiplication:** Multiply each component by a constant.

When combining these operations, as seen in the problem with \( 2\mathbf{v} - 3\mathbf{w} \), we handle each component separately to get the resultant vector \( \langle -5, 4 \rangle \). This illustrates how separate vectors can be combined to form a new vector.

Magnitude of a Vector

The magnitude of a vector gives us a measure of its length or size, regardless of its direction. It's calculated using the Pythagorean theorem. This notion applies to vectors in any dimension but is simple in two dimensions using the formula \( \sqrt{a^2 + b^2} \), where \( a \) and \( b \) are the components of the vector.
For the vector \( \langle -5, 4 \rangle \), its magnitude is \( \sqrt{(-5)^2 + 4^2} \). This simplifies to \( \sqrt{25 + 16} = \sqrt{41} \). Thus, the magnitude represents how long the vector is when stretched out straight.

• **Purpose of Magnitude:** Understanding the size of the vector.
• **Calculation:** Apply the Pythagorean theorem to the components.
• **Normalization Need:** You need the magnitude to normalize vectors effectively.

Grasping the concept of magnitude is crucial for further vector operations like normalization, which requires dividing by the magnitude to achieve a unit vector.

Component-wise Operations

Component-wise operations are straightforward yet powerful concepts in vector mathematics. These are operations performed on each component of the vector separately. Understanding this makes handling vector expressions seamless.

When we perform operations such as vector addition or subtraction, we do it component-wise. For example; given vectors \( \langle 4, 16 \rangle \) and \( \langle 9, 12 \rangle \), subtraction is performed like this:

• First component: \( 4 - 9 = -5 \)
• Second component: \( 16 - 12 = 4 \)

This results in the vector \( \langle -5, 4 \rangle \).
In scalar multiplication, each component is individually multiplied by the scalar. For a vector \( \langle a, b \rangle \) multiplied by a scalar \( c \), the result is \( \langle ac, bc \rangle \). This uniform manipulation across components ensures vectors retain their structure while scaling in size. Component-wise operations are foundational in tackling complex vector expressions by breaking them down into manageable parts.