Problem 20 - Thomas Calculus

Question

Sum of vectors \(\mathbf{u}=\mathbf{i}+(\mathbf{j}+\mathbf{k})\) is already the sum of a vector parallel to \(\mathbf{i}\) and a vector orthogonal to \(\mathbf{i}\) . If you use \(\mathbf{v}=\mathbf{i},\) in the decomposition \(\mathbf{u}=\operatorname{proj}_{v} \mathbf{u}+\left(\mathbf{u}-\operatorname{proj}_{v} \mathbf{u}\right),\) do you get projy \(\mathbf{u}=\mathbf{i}\) and \(\left(\mathbf{u}-\operatorname{proj}_{\mathbf{v}} \mathbf{u}\right)=\mathbf{j}+\mathbf{k} ?\) Try it and find out.

Step-by-Step Solution

Answer

Yes, \( \operatorname{proj}_{\mathbf{v}} \mathbf{u} = \mathbf{i} \) and \( \mathbf{u} - \operatorname{proj}_{\mathbf{v}} \mathbf{u} = \mathbf{j} + \mathbf{k} \).

Step 1: Understanding Vector Components

The given vector is \( \mathbf{u} = \mathbf{i} + (\mathbf{j} + \mathbf{k}) \), meaning it can already be viewed as the sum of two components: one along \( \mathbf{i} \) and one that is orthogonal to \( \mathbf{i} \).

Step 2: Determining Projection of \( \mathbf{u} \) on \( \mathbf{v} \)

Since \( \mathbf{v} = \mathbf{i} \), find the projection of \( \mathbf{u} \) on \( \mathbf{v} \) as \( \operatorname{proj}_{\mathbf{v}} \mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}} \mathbf{v} \).

Step 3: Calculate Dot Products

Calculate \( \mathbf{u} \cdot \mathbf{v} = (\mathbf{i} + \mathbf{j} + \mathbf{k}) \cdot \mathbf{i} = 1 \). Also, calculate \( \mathbf{v} \cdot \mathbf{v} = \mathbf{i} \cdot \mathbf{i} = 1 \).

Step 4: Find Projection

Substitute the dot products into the projection formula: \( \operatorname{proj}_{\mathbf{v}} \mathbf{u} = \frac{1}{1} \cdot \mathbf{i} = \mathbf{i} \).

Step 5: Determine Orthogonal Component

Find the orthogonal component as \( \mathbf{u} - \operatorname{proj}_{\mathbf{v}} \mathbf{u} = (\mathbf{i} + \mathbf{j} + \mathbf{k}) - \mathbf{i} = \mathbf{j} + \mathbf{k} \).

Step 6: Conclusion

Since \( \operatorname{proj}_{\mathbf{v}} \mathbf{u} = \mathbf{i} \) and \( \mathbf{u} - \operatorname{proj}_{\mathbf{v}} \mathbf{u} = \mathbf{j} + \mathbf{k} \), the proposed decomposition is correct.

Key Concepts

Projection of a Vector

When working with vectors, one common task is to find the way a vector can be projected onto another vector. This is called the projection of a vector. Essentially, this is finding how much of one vector goes in the direction of another. Visualize two vectors, \(\mathbf{u}\) and \(\mathbf{v}\). When we project \(\mathbf{u}\) onto \(\mathbf{v}\), we calculate a vector that is entirely in the direction of \(\mathbf{v}\).

Here’s how we calculate it: the formula for the projection of \(\mathbf{u}\) onto \(\mathbf{v}\) is \(\text{proj}_{\mathbf{v}} \mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}} \mathbf{v}\). This looks complicated, but let's break it down:

• The numerator, \(\mathbf{u} \cdot \mathbf{v}\), calculates how much the vectors "interact" with each other, essentially "how much" of \(\mathbf{u}\) is in the direction of \(\mathbf{v}\).
• The denominator, \(\mathbf{v} \cdot \mathbf{v}\), computes the scale of vector \(\mathbf{v}\).

In essence, this formula adjusts the direction of \(\mathbf{v}\) to fit the influence of \(\mathbf{u}\), giving us a new vector. In our original exercise, this projection turns out to be exactly \(\mathbf{i}\), as expected.

Orthogonal Component

Once you have the projection of one vector onto another, you'll often want to find what remains - that is, what part of the vector is orthogonal, or perpendicular, to the direction of the projection. This is known as the orthogonal component.

The orthogonal component is calculated by subtracting the projection from the original vector. Specifically, if you have \(\mathbf{u}\) and its projection \(\text{proj}_{\mathbf{v}} \mathbf{u}\), then the orthogonal component is \(\mathbf{u} - \text{proj}_{\mathbf{v}} \mathbf{u}\).

• This subtraction effectively removes the portion of \(\mathbf{u}\) that runs parallel to \(\mathbf{v}\).
• What you're left with is a vector that points entirely in directions orthogonal to \(\mathbf{v}\).

Following these steps in the problem, the orthogonal component of the vector was found to be \(\mathbf{j} + \mathbf{k}\), confirming that this part of vector \(\mathbf{u}\) does not align with \(\mathbf{i}\).

Dot Product Calculation

The dot product is a fundamental operation in vector mathematics and is especially important when calculating vector projections and orthogonal components. It helps measure how much two vectors align with each other, as well as fueling the projection formula we discussed earlier.

The dot product of two vectors \(\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}\) and \(\mathbf{b} = b_1\mathbf{i} + b_2\mathbf{j} + b_3\mathbf{k}\) is calculated as follows:
\[\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3\]

• The dot product is a scalar value, meaning it’s just a single number, not a vector.
• A high positive dot product can signal that two vectors point mostly in the same direction, while a dot product of zero indicates perpendicular vectors.

In the context of the original exercise, we calculated \(\mathbf{u} \cdot \mathbf{v}\) and \(\mathbf{v} \cdot \mathbf{v}\) to help pave the way for the vector projection. Each calculation tells us something about the directions of the vectors.