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Chapter 8: Problem 15

Find the magnitude of the projection of \(\langle 8,-4\rangle \) onto \(\langle 1,-3\rangle .\)

Short Answer

Expert verified
The magnitude of the projection is \(2\sqrt{10}\).

Step by step solution

01

Understand the Definition

The magnitude of the projection of a vector \( \mathbf{v} \) onto \( \mathbf{w} \) is given by the formula:\[\frac{|\mathbf{v} \cdot \mathbf{w}|}{\|\mathbf{w}\|}\]where \(\mathbf{v} = \langle v_1, v_2 \rangle\) and \(\mathbf{w} = \langle w_1, w_2 \rangle\).
02

Compute the Dot Product

The dot product \( \mathbf{v} \cdot \mathbf{w} \) is calculated as:\[v_1 \cdot w_1 + v_2 \cdot w_2\]For \( \langle 8, -4 \rangle \) and \( \langle 1, -3 \rangle \), it is:\[8 \times 1 + (-4) \times (-3) = 8 + 12 = 20\]
03

Find the Magnitude of \( \mathbf{w} \)

The magnitude of \( \mathbf{w} \), denoted \( \|\mathbf{w}\| \), is calculated using:\[\sqrt{w_1^2 + w_2^2}\]For \( \langle 1, -3 \rangle \), this becomes:\[\sqrt{1^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10}\]
04

Compute the Magnitude of the Projection

Use the values from the previous steps to find the magnitude of the projection:\[\frac{|\mathbf{v} \cdot \mathbf{w}|}{\|\mathbf{w}\|} = \frac{|20|}{\sqrt{10}} = \frac{20}{\sqrt{10}}\]Simplify \( \frac{20}{\sqrt{10}} \) by multiplying numerator and denominator by \( \sqrt{10} \) to rationalize the denominator:\[\frac{20\sqrt{10}}{10} = 2\sqrt{10}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Dot Product
The dot product is a fundamental operation in vector algebra. It's a way to multiply two vectors, resulting in a scalar. For two vectors \( \mathbf{v} = \langle v_1, v_2 \rangle \) and \( \mathbf{w} = \langle w_1, w_2 \rangle \), the dot product is found by multiplying their corresponding components and summing the results, given as:
  • \( \mathbf{v} \cdot \mathbf{w} = v_1 \cdot w_1 + v_2 \cdot w_2 \)
Let's look at an example, consider vectors \( \langle 8, -4 \rangle \) and \( \langle 1, -3 \rangle \). The dot product would be:
  • \( 8 \times 1 + (-4) \times (-3) = 8 + 12 = 20 \)
This operation combines the vectors into a single number, making it a useful measure in physics and computer graphics, where it's used to find angles between vectors or check if they are perpendicular.
Vector Magnitude
The magnitude of a vector represents its length or size, in the same way, a number defines the extent of a line segment. To compute the magnitude of a vector \( \mathbf{w} = \langle w_1, w_2 \rangle \), you use the Pythagorean Theorem:
  • \( \|\mathbf{w}\| = \sqrt{w_1^2 + w_2^2} \)
This provides a way to understand how large the vector is. For the vector \( \langle 1, -3 \rangle \), the magnitude calculation is:
  • \( \sqrt{1^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10} \)
The magnitude helps in various computations, such as normalizing a vector, which involves dividing each component by the magnitude to get a unit vector, maintaining its direction but with a length of one.
Rationalizing Denominators
Rationalizing the denominator refers to the process of eliminating any irrational numbers (square roots, for example) from the denominator of a fraction. This technique makes expressions simpler for further operations and easier to interpret. Let's demonstrate this with the fraction \( \frac{20}{\sqrt{10}} \).To rationalize, you multiply both the numerator and the denominator by the square root that appears in the denominator:
  • \( \frac{20}{\sqrt{10}} \cdot \frac{\sqrt{10}}{\sqrt{10}} = \frac{20\sqrt{10}}{10} \)
This operation removes the square root from the denominator, resulting in a more workable form:
  • \( 2\sqrt{10} \)
Using this process helps simplify expressions for clear communication and can be crucial in examinations and practical applications where precision is key.

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