https://file.aiquickdraw.com/mathbot/file/17294955508472r66heyy.jpg

To solve this problem, we need to find two components of vector **v** with respect to vector **b**:

1. The vector component of **v** along **b**, often referred to as the projection of **v** onto **b**.
2. The vector component of **v** orthogonal to **b**, which is the difference between **v** and its projection onto **b**.

### Step 1: Finding the Projection of **v** onto **b**

The formula for the projection of vector **v** onto vector **b** is:
\[
\text{proj}_b \mathbf{v} = \left( \frac{\mathbf{v} \cdot \mathbf{b}}{\mathbf{b} \cdot \mathbf{b}} \right) \mathbf{b}
\]
where:
- \(\mathbf{v} \cdot \mathbf{b}\) is the dot product of **v** and **b**.
- \(\mathbf{b} \cdot \mathbf{b}\) is the dot product of **b** with itself.

### Step 2: Finding the Orthogonal Component

The vector component of **v** orthogonal to **b** is given by:
\[
\mathbf{v}_{\perp} = \mathbf{v} - \text{proj}_b \mathbf{v}
\]

Let's solve both parts (a) and (b) using these steps.

### Part (a)

Given:
\[
\mathbf{v} = 2\hat{i} - \hat{j} + 3\hat{k}, \quad \mathbf{b} = \hat{i} + 2\hat{j} + 2\hat{k}
\]

1. **Calculate the dot product** \(\mathbf{v} \cdot \mathbf{b}\):
   \[
   \mathbf{v} \cdot \mathbf{b} = (2)(1) + (-1)(2) + (3)(2) = 2 - 2 + 6 = 6
   \]

2. **Calculate the dot product** \(\mathbf{b} \cdot \mathbf{b}\):
   \[
   \mathbf{b} \cdot \mathbf{b} = (1)^2 + (2)^2 + (2)^2 = 1 + 4 + 4 = 9
   \]

3. **Find the projection** \(\text{proj}_b \mathbf{v}\):
   \[
   \text{proj}_b \mathbf{v} = \left( \frac{6}{9} \right) \mathbf{b} = \frac{2}{3} \mathbf{b} = \frac{2}{3} (\hat{i} + 2\hat{j} + 2\hat{k}) = \left\langle \frac{2}{3}, \frac{4}{3}, \frac{4}{3} \right\rangle
   \]

4. **Find the orthogonal component** \(\mathbf{v}_{\perp}\):
   \[
   \mathbf{v}_{\perp} = \mathbf{v} - \text{proj}_b \mathbf{v} = \left\langle 2, -1, 3 \right\rangle - \left\langle \frac{2}{3}, \frac{4}{3}, \frac{4}{3} \right\rangle = \left\langle \frac{4}{3}, -\frac{7}{3}, \frac{5}{3} \right\rangle
   \]

### Part (b)

Given:
\[
\mathbf{v} = \langle 4, -1, 7 \rangle, \quad \mathbf{b} = \langle 2, 3, -6 \rangle
\]

1. **Calculate the dot product** \(\mathbf{v} \cdot \mathbf{b}\):
   \[
   \mathbf{v} \cdot \mathbf{b} = (4)(2) + (-1)(3) + (7)(-6) = 8 - 3 - 42 = -37
   \]

2. **Calculate the dot product** \(\mathbf{b} \cdot \mathbf{b}\):
   \[
   \mathbf{b} \cdot \mathbf{b} = (2)^2 + (3)^2 + (-6)^2 = 4 + 9 + 36 = 49
   \]

3. **Find the projection** \(\text{proj}_b \mathbf{v}\):
   \[
   \text{proj}_b \mathbf{v} = \left( \frac{-37}{49} \right) \mathbf{b} = \frac{-37}{49} \langle 2, 3, -6 \rangle = \left\langle \frac{-74}{49}, \frac{-111}{49}, \frac{222}{49} \right\rangle
   \]

4. **Find the orthogonal component** \(\mathbf{v}_{\perp}\):
   \[
   \mathbf{v}_{\perp} = \mathbf{v} - \text{proj}_b \mathbf{v} = \langle 4, -1, 7 \rangle - \left\langle \frac{-74}{49}, \frac{-111}{49}, \frac{222}{49} \right\rangle = \left\langle \frac{270}{49}, \frac{62}{49}, \frac{121}{49} \right\rangle
   \]

Would you like more details on the calculations or have any questions?

### Related Questions:
1. How do you calculate the dot product of two vectors?
2. What is the significance of the projection of a vector onto another vector?
3. How can you determine if two vectors are orthogonal?
4. What is the geometric interpretation of the orthogonal component of a vector?
5. How can vector components be used in physics for resolving forces?

### Tip:
When finding the projection of one vector onto another, make sure the vector being projected onto (in this case, **b**) is not the zero vector, as division by zero would be undefined.

In each part, find the vector component of v along b and the vector component of v orthogonal to b. (a) v = 2i - j + 3k, b = i + 2j + 2k; (b) v = <4, -1, 7>, b = <2, 3, -6>.

Solve the problem of finding the vector component of v along b and the orthogonal component of v to b. Learn to calculate projections using the dot product formula for vector decomposition, including both projection and orthogonal components. Suitable for advanced high school and undergraduate students.

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