In each part, find the vector component of ~v along ~b and the vector component of ~v orthogonal to ~b. Then sketch the vectors ~v, proj~b ~v, and ~v − proj~b ~v. (a) ~v = 2ˆi −ˆj, ~b = 3ˆi +ˆj

We are tasked with finding two vector components of \( \vec{v} \) with respect to \( \vec{b} \):
1. The vector component of \( \vec{v} \) along \( \vec{b} \) (denoted as \( \text{proj}_{\vec{b}} \vec{v} \)).
2. The vector component of \( \vec{v} \) orthogonal to \( \vec{b} \), which is \( \vec{v} - \text{proj}_{\vec{b}} \vec{v} \).

### Given:
\[
\vec{v} = 2\hat{i} - \hat{j} = \begin{pmatrix} 2 \\ -1 \end{pmatrix}, \quad \vec{b} = 3\hat{i} + \hat{j} = \begin{pmatrix} 3 \\ 1 \end{pmatrix}
\]

### Step 1: Calculate the projection of \( \vec{v} \) along \( \vec{b} \)
The formula for the projection of a vector \( \vec{v} \) along another vector \( \vec{b} \) is:
\[
\text{proj}_{\vec{b}} \vec{v} = \frac{\vec{v} \cdot \vec{b}}{\vec{b} \cdot \vec{b}} \vec{b}
\]

#### Find \( \vec{v} \cdot \vec{b} \) (dot product):
\[
\vec{v} \cdot \vec{b} = (2)(3) + (-1)(1) = 6 - 1 = 5
\]

#### Find \( \vec{b} \cdot \vec{b} \):
\[
\vec{b} \cdot \vec{b} = (3)^2 + (1)^2 = 9 + 1 = 10
\]

#### Now, calculate \( \text{proj}_{\vec{b}} \vec{v} \):
\[
\text{proj}_{\vec{b}} \vec{v} = \frac{5}{10} \vec{b} = \frac{1}{2} \begin{pmatrix} 3 \\ 1 \end{pmatrix} = \begin{pmatrix} \frac{3}{2} \\ \frac{1}{2} \end{pmatrix}
\]

### Step 2: Calculate the vector component of \( \vec{v} \) orthogonal to \( \vec{b} \)
This component is simply \( \vec{v} - \text{proj}_{\vec{b}} \vec{v} \):
\[
\vec{v} - \text{proj}_{\vec{b}} \vec{v} = \begin{pmatrix} 2 \\ -1 \end{pmatrix} - \begin{pmatrix} \frac{3}{2} \\ \frac{1}{2} \end{pmatrix} = \begin{pmatrix} 2 - \frac{3}{2} \\ -1 - \frac{1}{2} \end{pmatrix} = \begin{pmatrix} \frac{1}{2} \\ -\frac{3}{2} \end{pmatrix}
\]

### Step 3: Sketch the vectors

The vectors to be sketched are:
- \( \vec{v} = \begin{pmatrix} 2 \\ -1 \end{pmatrix} \)
- \( \text{proj}_{\vec{b}} \vec{v} = \begin{pmatrix} \frac{3}{2} \\ \frac{1}{2} \end{pmatrix} \)
- \( \vec{v} - \text{proj}_{\vec{b}} \vec{v} = \begin{pmatrix} \frac{1}{2} \\ -\frac{3}{2} \end{pmatrix} \)

I'll prepare the sketch, which shows \( \vec{v} \), \( \text{proj}_{\vec{b}} \vec{v} \), and the orthogonal component \( \vec{v} - \text{proj}_{\vec{b}} \vec{v} \).

Would you like a more detailed explanation of the process? 

Meanwhile, here are 5 questions to expand on the topic:
1. How is the projection of a vector geometrically interpreted?
2. What is the significance of the dot product in determining the projection?
3. Can you explain why \( \vec{v} - \text{proj}_{\vec{b}} \vec{v} \) is orthogonal to \( \vec{b} \)?
4. How does the projection formula change when vectors are not in standard form?
5. What are some real-world applications of vector projection?

**Tip**: The vector projection represents the "shadow" of one vector onto another, giving the component that aligns with the second vector.

This problem explores how to find the vector component of ~v along ~b and the vector component orthogonal to ~b for ~v = 2i - j and ~b = 3i + j. It covers concepts like vector projection, dot products, and orthogonality. Suitable for Grades 10-12 students.

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