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.