Problem 18 - Thomas Calculus

Question

In Exercises \(17-19\) , write \(u\) as the sum of a vector parallel to \(v\) and a vector orthogonal to \(v\) . $$ \mathbf{u}=\mathbf{j}+\mathbf{k}, \quad \mathbf{v}=\mathbf{i}+\mathbf{j} $$

Step-by-Step Solution

Answer

\(\mathbf{u} = \left(\frac{1}{2} \mathbf{i} + \frac{1}{2} \mathbf{j}\right) + \left(-\frac{1}{2} \mathbf{i} + \frac{1}{2} \mathbf{j} + \mathbf{k}\right)\).

Step 1: Find the projection of \(u\) onto \(v\)

To find the vector that is parallel to \(v\), we need to compute the projection of \(u\) onto \(v\). The formula for the projection of \(\mathbf{u}\) onto \(\mathbf{v}\) is \( \text{proj}_{\mathbf{v}} \mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}} \mathbf{v} \). First, compute the dot products \(\mathbf{u} \cdot \mathbf{v}\) and \(\mathbf{v} \cdot \mathbf{v}\).

Step 2: Compute the dot products

Calculate \(\mathbf{u} \cdot \mathbf{v}\) by expanding it as \((0\cdot1) + (1\cdot1) + (1\cdot0) = 1\). Compute \(\mathbf{v} \cdot \mathbf{v}\) as \((1\cdot1) + (1\cdot1) + (0\cdot0) = 2\).

Step 3: Find the projection vector

Substitute the dot product results back into the projection formula: \(\text{proj}_{\mathbf{v}} \mathbf{u} = \frac{1}{2} (\mathbf{i} + \mathbf{j}) = \frac{1}{2} \mathbf{i} + \frac{1}{2} \mathbf{j}\). So, the projection of \(\mathbf{u}\) onto \(\mathbf{v}\) is \(\frac{1}{2} \mathbf{i} + \frac{1}{2} \mathbf{j}\).

Step 4: Subtract to find the orthogonal component

To find the orthogonal vector, subtract the projection from \(\mathbf{u}\). Compute \(\mathbf{u} - \text{proj}_{\mathbf{v}} \mathbf{u} = (0\mathbf{i} + 1\mathbf{j} + 1\mathbf{k}) - \left(\frac{1}{2} \mathbf{i} + \frac{1}{2} \mathbf{j}\right) = -\frac{1}{2} \mathbf{i} + \frac{1}{2} \mathbf{j} + \mathbf{k}\).

Step 5: Write \(u\) as a sum of the parallel and orthogonal vectors

Now that we have both components, we can express \(\mathbf{u}\) as the sum of the vector parallel to \(\mathbf{v}\) and the vector orthogonal to \(\mathbf{v}\): \(\mathbf{u} = \left(\frac{1}{2} \mathbf{i} + \frac{1}{2} \mathbf{j}\right) + \left(-\frac{1}{2} \mathbf{i} + \frac{1}{2} \mathbf{j} + \mathbf{k}\right)\).

Key Concepts

Vector Projection

The concept of vector projection is essential when splitting vectors into components that are parallel and orthogonal. The projection of one vector onto another is a way of finding the closest point on a vector line to another vector. It helps in creating a component of a vector that lies in the same direction as another vector.

To compute the projection of vector \( \mathbf{u} \) onto vector \( \mathbf{v} \), we use the formula:

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

The formula uses dot products to determine the scale of vector \( \mathbf{v} \) needed to best represent \( \mathbf{u} \) along \( \mathbf{v} \). The fraction represents how much of \( \mathbf{u} \) can be attributed to \( \mathbf{v} \).
This process plays a critical role when decomposing vectors into parallel and orthogonal components during vector decomposition.

Dot Product

The dot product, also known as scalar product, is fundamental in vector mathematics. It measures how much of one vector goes in the direction of another.

For any two vectors \( \mathbf{a} = a_1 \mathbf{i} + a_2 \mathbf{j} + a_3 \mathbf{k} \) and \( \mathbf{b} = b_1 \mathbf{i} + b_2 \mathbf{j} + b_3 \mathbf{k} \), the dot product \( \mathbf{a} \cdot \mathbf{b} \) is calculated as:

• \( a_1b_1 + a_2b_2 + a_3b_3 \)

This value is a scalar, indicating the directional alignment of the two vectors. When calculating projections, the dot product enables us to find how much of one vector is present in another.

In the case of \( \mathbf{u} \) and \( \mathbf{v} \) from the exercise, computing \( \mathbf{u} \cdot \mathbf{v} \) provides a scalar value needed for determining the projection.
Moreover, the dot product is also pivotal in checking the orthogonality of vectors, where it results in zero for orthogonal vectors.

Orthogonal Vectors

Orthogonal vectors are vectors that are perpendicular to each other. This relationship is crucial in vector decomposition, aiding in separating components effectively. When two vectors are orthogonal, their dot product is zero, meaning they have no directional overlap.

Finding the orthogonal component of a vector involves subtracting the projection from the original vector:

• Given \( \mathbf{u} \) and its projection \( \text{proj}_{\mathbf{v}} \mathbf{u} \), the orthogonal projection is \( \mathbf{u} - \text{proj}_{\mathbf{v}} \mathbf{u} \).

This process leaves us with a vector that is completely independent directionally from \( \mathbf{v} \).
In the exercise example, after obtaining the projection, the remaining part of \( \mathbf{u} \) represents the vector orthogonal to \( \mathbf{v} \), decomposing \( \mathbf{u} \) into its parallel and orthogonal components.