Problem 25 - Algebra and Trigonometry

Question

Find the component of \(\mathbf{u}\) along \(\mathbf{v}\) $$ \mathbf{u}=\langle 4,6\rangle, \quad \mathbf{v}=\langle 3,-4\rangle $$

Step-by-Step Solution

Answer

The component of \( \mathbf{u} \) along \( \mathbf{v} \) is \(-2.4\).

Step 1: Find the Dot Product

Calculate the dot product of the vectors \( \mathbf{u} \) and \( \mathbf{v} \). The formula for the dot product is: \( \mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 \). Here, \( \mathbf{u} = \langle 4, 6 \rangle \) and \( \mathbf{v} = \langle 3, -4 \rangle \). Substitute the values: \( 4 \times 3 + 6 \times (-4) = 12 - 24 = -12 \). The dot product is \( -12 \).

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

Calculate the magnitude of \( \mathbf{v} \). The formula for the magnitude is: \( \| \mathbf{v} \| = \sqrt{v_1^2 + v_2^2} \). Substitute the values: \( \| \mathbf{v} \| = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \). So, the magnitude of \( \mathbf{v} \) is \( 5 \).

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

The component of \( \mathbf{u} \) along \( \mathbf{v} \) is given by the formula: \( \text{comp}_\mathbf{v}(\mathbf{u}) = \frac{\mathbf{u} \cdot \mathbf{v}}{\| \mathbf{v} \|} \). Substitute the values: \( \text{comp}_\mathbf{v}(\mathbf{u}) = \frac{-12}{5} = -2.4 \). Thus, the component of \( \mathbf{u} \) along \( \mathbf{v} \) is \( -2.4 \).

Key Concepts

Dot Product

The dot product, also known as the scalar product, is a fundamental operation in vector mathematics. It's used to find the relationship between two vectors in terms of direction and magnitude. The dot product of two vectors \( \mathbf{u} = \langle u_1, u_2 \rangle \) and \( \mathbf{v} = \langle v_1, v_2 \rangle \) is calculated as:

• \( \mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 \)

In our exercise, we're given the vectors \( \mathbf{u} = \langle 4, 6 \rangle \) and \( \mathbf{v} = \langle 3, -4 \rangle \). By substituting, we find the dot product:

• \( 4 \times 3 + 6 \times (-4) = 12 - 24 = -12 \).

The result, \(-12\), is a scalar quantity that reveals how much \( \mathbf{u} \) aligns with \( \mathbf{v} \) but does not explicitly reveal the direction.

Magnitude of a Vector

The magnitude of a vector, often referred to as its length or norm, is a measure of how long the vector is. It is calculated using the Pythagorean theorem for a vector \( \mathbf{v} = \langle v_1, v_2 \rangle \). The formula to find the magnitude is:

• \( \| \mathbf{v} \| = \sqrt{v_1^2 + v_2^2} \)

For vector \( \mathbf{v} = \langle 3, -4 \rangle \), the magnitude is calculated by substituting the values:

• \( \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \).

This tells us that the length of \( \mathbf{v} \) is 5 units, essential for understanding how the vector stretches in its direction.

Component of a Vector

The component of vector \( \mathbf{u} \) along another vector \( \mathbf{v} \) is a representation of \( \mathbf{u} \) projected onto \( \mathbf{v} \). This helps us understand how much of \( \mathbf{u} \) lies in the direction of \( \mathbf{v} \). The formula used is:

• \( \text{comp}_\mathbf{v}(\mathbf{u}) = \frac{\mathbf{u} \cdot \mathbf{v}}{\| \mathbf{v} \|} \)

For the given vectors, where \( \mathbf{u} \cdot \mathbf{v} = -12 \) and \( \| \mathbf{v} \| = 5 \), we find:

• \( \text{comp}_\mathbf{v}(\mathbf{u}) = \frac{-12}{5} = -2.4 \)

The component \(-2.4\) indicates how much of \( \mathbf{u} \) runs parallel to \( \mathbf{v} \); here, negative means it's in the opposite direction of \( \mathbf{v} \). This is a crucial concept in vector projection that shapes our understanding of vector interactions in space.