Problem 17 - Thomas Calculus

Question

In Exercises \(17-19\) , write \(u\) as the sum of a vector parallel to \(v\) and a vector orthogonal to \(v\) . $$ \mathbf{u}=3 \mathbf{j}+4 \mathbf{k}, \quad \mathbf{v}=\mathbf{i}+\mathbf{j} $$

Step-by-Step Solution

Answer

\(\mathbf{u}\) is \(\frac{3}{2}\mathbf{i} + \frac{3}{2}\mathbf{j}\) (parallel) + \(-\frac{3}{2}\mathbf{i} + \frac{3}{2}\mathbf{j} + 4\mathbf{k}\) (orthogonal).

Step 1: Find the Projection of u onto v

The projection of vector \(\mathbf{u}\) onto vector \(\mathbf{v}\) can be calculated using the formula: \[ \text{proj}_{\mathbf{v}} \mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}} \mathbf{v} \] First, compute the dot product \(\mathbf{u} \cdot \mathbf{v}\): \[ \mathbf{u} \cdot \mathbf{v} = (3\mathbf{j}+4\mathbf{k}) \cdot (\mathbf{i} + \mathbf{j}) = 0 + 3 = 3 \] Next, compute \(\mathbf{v} \cdot \mathbf{v}\): \[ \mathbf{v} \cdot \mathbf{v} = (\mathbf{i} + \mathbf{j}) \cdot (\mathbf{i} + \mathbf{j}) = 1 + 1 = 2 \] Now, substitute these values back into the formula: \[ \text{proj}_{\mathbf{v}} \mathbf{u} = \frac{3}{2}(\mathbf{i} + \mathbf{j}) = \frac{3}{2}\mathbf{i} + \frac{3}{2}\mathbf{j} \]

Step 2: Find the Orthogonal Component

The vector \(\mathbf{u}\) can be expressed as the sum of a vector parallel to \(\mathbf{v}\) (which we found in Step 1) and a vector orthogonal to \(\mathbf{v})\). This orthogonal component \(\mathbf{u}_{\perp}\) is given by:\[ \mathbf{u}_{\perp} = \mathbf{u} - \text{proj}_{\mathbf{v}} \mathbf{u} \] Substitute to find \(\mathbf{u}_{\perp}\): \[ \mathbf{u}_{\perp} = (3\mathbf{j} + 4\mathbf{k}) - \left(\frac{3}{2}\mathbf{i} + \frac{3}{2}\mathbf{j}\right) \]Distribute and simplify:\[ \mathbf{u}_{\perp} = 0\mathbf{i} + 3\mathbf{j} + 4\mathbf{k} - \frac{3}{2}\mathbf{i} - \frac{3}{2}\mathbf{j} = -\frac{3}{2}\mathbf{i} + \frac{3}{2}\mathbf{j} + 4\mathbf{k} \]

Step 3: Summarize the Result

Summarize the components of \(\mathbf{u}\):The vector \(\mathbf{u}\) can be written as the sum of a vector parallel to \(\mathbf{v}\) and a vector orthogonal to \(\mathbf{v}\):\[ \mathbf{u} = \left( \frac{3}{2} \mathbf{i} + \frac{3}{2} \mathbf{j} \right) + \left( -\frac{3}{2}\mathbf{i} + \frac{3}{2}\mathbf{j} + 4\mathbf{k} \right) \] The components are:- Parallel to \(\mathbf{v}\): \( \frac{3}{2} \mathbf{i} + \frac{3}{2} \mathbf{j} \)- Orthogonal to \(\mathbf{v}\): \(-\frac{3}{2}\mathbf{i} + \frac{3}{2}\mathbf{j} + 4\mathbf{k} \)

Key Concepts

Projection of Vectors

Vector projection helps us find a component of a vector that points in the same direction as another vector. Imagine a flashlight is shining on a vector from above, casting a shadow onto another vector, this shadow is similar to the projection. To calculate the projection of vector \( \mathbf{u} \) onto vector \( \mathbf{v} \), we use the formula:

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

This formula uses the dot product between \( \mathbf{u} \) and \( \mathbf{v} \). It expresses how much of \( \mathbf{u} \) aligns with \( \mathbf{v} \). The result is a vector that is parallel to \( \mathbf{v} \). Once we calculate the dot products involved and substitute into the formula, it determines the parallel component of \( \mathbf{u} \) to \( \mathbf{v} \). This is crucial in breaking down vectors for analysis into directions of interest, particularly in physics and engineering applications.

Orthogonal Vectors

Orthogonal vectors are vectors that intersect at right angles, like streets crossing each other at a 90-degree angle. The orthogonal component of a vector relative to another vector is significant in many applications, such as minimizing distances or resolving forces in directions not aligned with each other.
To find an orthogonal vector or component, we subtract the projection vector from the original vector. So, if you have vector \( \mathbf{u} \), and you've already found \( \text{proj}_{\mathbf{v}} \mathbf{u} \), the orthogonal component \( \mathbf{u}_{\perp} \) is:

• \( \mathbf{u}_{\perp} = \mathbf{u} - \text{proj}_{\mathbf{v}} \mathbf{u} \)

By doing this, you find what part of \( \mathbf{u} \) does not lie along \( \mathbf{v} \). Essentially, it leaves you with a vector that sits entirely outside the direction defined by \( \mathbf{v} \). Understanding orthogonality is essential for vector spaces and can be applied across disciplines such as signal processing, computer graphics, and more.

Dot Product Calculation

The dot product is a way of multiplying two vectors that tells us how much one vector extends in the direction of another. Instead of multiplying the vectors' magnitudes directly like in scalars, the dot product considers their direction too. The formula for the dot product of two vectors \( \mathbf{u} = a_1\mathbf{i} + b_1\mathbf{j} + c_1\mathbf{k} \) and \( \mathbf{v} = a_2\mathbf{i} + b_2\mathbf{j} + c_2\mathbf{k} \) is:

• \( \mathbf{u} \cdot \mathbf{v} = a_1a_2 + b_1b_2 + c_1c_2 \)

This process sums up the products of corresponding components of the vectors. If the dot product is zero, the vectors are orthogonal, meaning they are at right angles to each other.
Dot products are used to determine angles between vectors, for calculating work done by a force, and setting up projections. Additionally, it inflates into more complex calculations in fields like quantum mechanics and computer graphics, making it a foundational concept in vector math.