Problem 28 - Calculus Early Transcendentals: Pearson New International Edition

Question

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 \(\mathbf{w}=\mathbf{i}+5 \mathbf{j}\). $$ \operatorname{proj}_{\mathbf{i}} \mathbf{u} $$

Step-by-Step Solution

Answer

The projection of \( \mathbf{u} \) onto \( \mathbf{i} \) is \( \mathbf{i} \).

Step 1: 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 \( \mathbf{i} \).

Step 2: 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: \[ \mathbf{u} \cdot \mathbf{i} = (1)(1) + (2)(0) = 1 \]

Step 3: 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: \[ \mathbf{i} \cdot \mathbf{i} = (1)(1) + (0)(0) = 1 \]

Step 4: Compute the Projection

Use the results from Steps 2 and 3 to find the projection of \( \mathbf{u} \) onto \( \mathbf{i} \): \[ \operatorname{proj}_{\mathbf{i}} \mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{i}}{\mathbf{i} \cdot \mathbf{i}} \mathbf{i} = \frac{1}{1} \mathbf{i} = \mathbf{i} \]

Step 5: Conclusion

The projection of \( \mathbf{u} \) onto \( \mathbf{i} \) is simply \( \mathbf{i} \), as calculated in Step 4.

Key Concepts

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 \(a_{x}b_{x} + a_{y}b_{y} + a_{z}b_{z}\).

The beauty of the dot product is that it is zero when the vectors are perpendicular, revealing the angle between them. And, when used in the projection formula, it helps us determine how much of one vector projects onto another. In the original exercise, we computed the dot product \(\mathbf{u} \cdot \mathbf{i} = 1\). This tells us how much of the vector \(\mathbf{u}\) goes in the direction of \(\mathbf{i}\).

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:

• 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.

By understanding vector components, we can better grasp how vectors operate in different dimensions. Components are essential for computing dot products, as they help break complex equations into manageable parts.
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 \(\mathbf{b}\).

In the case of the exercise, when we projected \(\mathbf{u}\) onto \(\mathbf{i}\), we demonstrated the application of this formula: The projection \(\operatorname{proj}_{\mathbf{i}} \mathbf{u}\) equaled \(\mathbf{i}\), showing vector \(\mathbf{u}\)'s effectiveness in the direction of \(\mathbf{i}\). This technique is very practical, not just in mathematics but also in physics and engineering!