Chapter 11: Problem 28
In Problems 23-28, find each of the given projections if \(\mathbf{u}=\mathbf{i}+2 \mathbf{j}, \mathbf{v}=2 \mathbf{i}-\mathbf{j}\), and \(\mathbf{w}=\mathbf{i}+5 \mathbf{j}\). $$ \operatorname{proj}_{\mathbf{i}} \mathbf{u} $$
Short Answer
Expert verified
The projection of \( \mathbf{u} \) onto \( \mathbf{i} \) is \( \mathbf{i} \).
Step by step solution
01
Recall the Projection Formula
The projection of a vector \( \mathbf{a} \) onto another vector \( \mathbf{b} \) is given by the formula: \[ \operatorname{proj}_{\mathbf{b}} \mathbf{a} = \frac{\mathbf{a} \cdot \mathbf{b}}{\mathbf{b} \cdot \mathbf{b}} \mathbf{b} \] In this case, \( \mathbf{a} = \mathbf{u} \) and \( \mathbf{b} = \mathbf{i} \). We will use this formula to find the projection of \( \mathbf{u} \) onto \( \mathbf{i} \).
02
Find the Dot Product \(\mathbf{u} \cdot \mathbf{i}\)
Compute the dot product \( \mathbf{u} \cdot \mathbf{i} \). The vector \( \mathbf{u} = \mathbf{i} + 2 \mathbf{j} \) and \( \mathbf{i} = 1\mathbf{i} + 0\mathbf{j} \). The dot product is computed as: \[ \mathbf{u} \cdot \mathbf{i} = (1)(1) + (2)(0) = 1 \]
03
Find \(\mathbf{i} \cdot \mathbf{i}\)
Compute the dot product \( \mathbf{i} \cdot \mathbf{i} \). Since \( \mathbf{i} = 1\mathbf{i} + 0\mathbf{j} \), the dot product is: \[ \mathbf{i} \cdot \mathbf{i} = (1)(1) + (0)(0) = 1 \]
04
Compute the Projection
Use the results from Steps 2 and 3 to find the projection of \( \mathbf{u} \) onto \( \mathbf{i} \): \[ \operatorname{proj}_{\mathbf{i}} \mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{i}}{\mathbf{i} \cdot \mathbf{i}} \mathbf{i} = \frac{1}{1} \mathbf{i} = \mathbf{i} \]
05
Conclusion
The projection of \( \mathbf{u} \) onto \( \mathbf{i} \) is simply \( \mathbf{i} \), as calculated in Step 4.
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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 when dealing with vectors. Imagine two vectors, \(\mathbf{a}\) and \(\mathbf{b}\). The dot product of these vectors is a way to multiply them to get a scalar, or a number, rather than another vector. This is handy because it helps us understand the relationship between the two vectors, including angles and projections.
- For two-dimensional vectors, the dot product is calculated as \(\mathbf{a} \cdot \mathbf{b} = a_{x}b_{x} + a_{y}b_{y}\).
- If the vectors are in three dimensions, it's \(a_{x}b_{x} + a_{y}b_{y} + a_{z}b_{z}\).
Vector Components
A vector can be thought of as a kind of arrow showing direction and magnitude. But when directions vary in space, how do we break down a vector? This is where components come in handy.
Components help us see a vector as a sum of individual parts that point along different axes. For instance, in a 2D space:
In the exercise, identifying \(\mathbf{i}\)'s components were crucial for finding \(\mathbf{u} \cdot \mathbf{i}\).
Components help us see a vector as a sum of individual parts that point along different axes. For instance, in a 2D space:
- A vector \(\mathbf{u} = \mathbf{i} + 2 \mathbf{j}\) means its components are parallel to the x-axis (\(\mathbf{i}\)) and y-axis (\(\mathbf{j}\)).
- \(\mathbf{i}\) has components \(1\mathbf{i} + 0\mathbf{j}\), pointing only in the x-direction.
In the exercise, identifying \(\mathbf{i}\)'s components were crucial for finding \(\mathbf{u} \cdot \mathbf{i}\).
Projection Formula
Projecting one vector onto another is like casting a shadow; it tells us how much of one vector falls along the direction of another. We often use the projection formula:\[\operatorname{proj}_{\mathbf{b}} \mathbf{a} = \frac{\mathbf{a} \cdot \mathbf{b}}{\mathbf{b} \cdot \mathbf{b}} \mathbf{b}\]This formula helps express the piece of vector \(\mathbf{a}\) that aligns with vector \(\mathbf{b}\). Here’s how it works:
- Find the dot product of your two vectors. This gives a scalar that reflects their directional similarity.
- Divide by the dot product of vector \(\mathbf{b}\) itself. This ensures the projection’s scaling is accurate.
- The result is a vector projecting \(\mathbf{a}\) onto \(\mathbf{b}\).