Problem 26 - Precalculus Mathematics for Calculus

Question

Find the component of \(\mathbf{u}\) along \(\mathbf{v}\). $$\mathbf{u}=\langle- 3,5\rangle, \quad \mathbf{v}=\langle 1 / \sqrt{2}, 1 / \sqrt{2}\rangle$$

Step-by-Step Solution

Answer

The component is \(\sqrt{2}\).

Step 1: Compute the Dot Product

First, calculate the dot product of \(\mathbf{u}\) and \(\mathbf{v}\). The dot product is given by \( \mathbf{u} \cdot \mathbf{v} = -3 \cdot \frac{1}{\sqrt{2}} + 5 \cdot \frac{1}{\sqrt{2}} = \frac{-3}{\sqrt{2}} + \frac{5}{\sqrt{2}} = \frac{2}{\sqrt{2}} \).

Step 2: Find Magnitude of \(\mathbf{v}\)

The magnitude of a vector \(\mathbf{v} = \langle a, b \rangle\) is calculated using \( \| \mathbf{v} \| = \sqrt{a^2 + b^2} \). For \(\mathbf{v} = \langle \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \rangle \), this becomes \( \| \mathbf{v} \| = \sqrt{\left( \frac{1}{\sqrt{2}} \right)^2 + \left( \frac{1}{\sqrt{2}} \right)^2} = \sqrt{\frac{1}{2} + \frac{1}{2}} = 1 \).

Step 3: Calculate the Component of \(\mathbf{u}\) along \(\mathbf{v}\)

The component of \(\mathbf{u}\) along \(\mathbf{v}\) is given by \( \text{comp}_{\mathbf{v}} \mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{v}}{\| \mathbf{v} \|} \). Since \(\| \mathbf{v} \| = 1\), this simplifies to \( \text{comp}_{\mathbf{v}} \mathbf{u} = \frac{2}{\sqrt{2}} = \sqrt{2} \).

Key Concepts

Dot Product

The dot product is a way to multiply two vectors, resulting in a scalar (a single number). It helps measure how much one vector goes in the direction of another. The dot product of two vectors \( \mathbf{u} \) and \( \mathbf{v} \) can be calculated with the formula:
\[ \mathbf{u} \cdot \mathbf{v} = u_1 \cdot v_1 + u_2 \cdot v_2 + \ldots + u_n \cdot v_n \]
Here, \( u_i \) and \( v_i \) are the components of the vectors. For the vectors \( \mathbf{u} = \langle -3, 5 \rangle \) and \( \mathbf{v} = \langle \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \rangle \), we calculate the dot product as follows:

• Multiply corresponding components: \(-3 \times \frac{1}{\sqrt{2}}\) and \(5 \times \frac{1}{\sqrt{2}}\).
• Add these products: \(\frac{-3}{\sqrt{2}} + \frac{5}{\sqrt{2}}\).

Simplifying gives the result \(\frac{2}{\sqrt{2}}\). The dot product highlights the extent to which \( \mathbf{u} \) aligns with \( \mathbf{v} \).

Vector Magnitude

Vector magnitude signifies the length or size of a vector. It's like finding the distance of the vector from the origin. For any vector \( \mathbf{v} = \langle a, b \rangle \), the magnitude is calculated using the Pythagorean theorem:
\[ \| \mathbf{v} \| = \sqrt{a^2 + b^2} \]
The vector \( \mathbf{v} = \langle \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \rangle \) presents an interesting scenario:

• Square each component: \(\left(\frac{1}{\sqrt{2}}\right) ^2 = \frac{1}{2}\).
• Add them up: \(\frac{1}{2} + \frac{1}{2} = 1\).
• Take the square root: \(\sqrt{1} = 1\).

This results in a magnitude of \(1\), indicating that \( \mathbf{v} \) is a unit vector, having a length of 1.

Vector Projection

Vector projection involves identifying how much of one vector lies in the direction of another. It is like casting one vector onto another to see what portion aligns with it. The component, or projection, of vector \( \mathbf{u} \) along vector \( \mathbf{v} \) is given by:
\[ \text{comp}_{\mathbf{v}} \mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{v}}{\| \mathbf{v} \|} \]
Using our example:

• We already know the dot product: \(\frac{2}{\sqrt{2}}\).
• The magnitude of \( \mathbf{v} \) is \(1\).

By dividing the dot product by the magnitude, the formula simplifies to:\[\text{comp}_{\mathbf{v}} \mathbf{u} = \frac{2}{\sqrt{2}} = \sqrt{2}\]This result tells us how much of \( \mathbf{u} \) points in the direction of \( \mathbf{v} \), essentially measuring its effectiveness when overlayed onto \( \mathbf{v} \).