The problem involves determining the component of vector a that is perpendicular to vector b. Here's how to solve it step by step:

Step 1: Understand the Vectors

• a = (3√2, 3√3, 3)
• b = (√2, -√3, 2)

Step 2: Orthogonal (Perpendicular) Component Formula

The component of vector a that is perpendicular to b can be found by subtracting the projection of a onto b from a:

$\text{Perpendicular component of } \mathbf{a} = \mathbf{a} - \text{Proj}_{\mathbf{b}} \mathbf{a}$

Where the projection of a onto b is given by:

$\text{Proj}_{\mathbf{b}} \mathbf{a} = \frac{\mathbf{a} \cdot \mathbf{b}}{\mathbf{b} \cdot \mathbf{b}} \mathbf{b}$

Step 3: Compute the Dot Products

1. Dot product of a and b ($\mathbf{a} \cdot \mathbf{b}$):

$\mathbf{a} \cdot \mathbf{b} = (3\sqrt{2})(\sqrt{2}) + (3\sqrt{3})(-\sqrt{3}) + (3)(2)$

$\mathbf{a} \cdot \mathbf{b} = 6 - 9 + 6 = 3$

2. Dot product of b and b ($\mathbf{b} \cdot \mathbf{b}$):

$\mathbf{b} \cdot \mathbf{b} = (\sqrt{2})^2 + (-\sqrt{3})^2 + (2)^2$

$\mathbf{b} \cdot \mathbf{b} = 2 + 3 + 4 = 9$

Step 4: Compute the Projection of a onto b

The scalar projection factor is:

$\frac{\mathbf{a} \cdot \mathbf{b}}{\mathbf{b} \cdot \mathbf{b}} = \frac{3}{9} = \frac{1}{3}$

Thus, the projection vector is:

$\text{Proj}_{\mathbf{b}} \mathbf{a} = \frac{1}{3} \mathbf{b} = \frac{1}{3} (\sqrt{2}, -\sqrt{3}, 2)$

$\text{Proj}_{\mathbf{b}} \mathbf{a} = \left(\frac{\sqrt{2}}{3}, -\frac{\sqrt{3}}{3}, \frac{2}{3}\right)$

Step 5: Compute the Perpendicular Component

Subtract the projection vector from a:

$\text{Perpendicular component of } \mathbf{a} = \mathbf{a} - \text{Proj}_{\mathbf{b}} \mathbf{a}$

$\text{Perpendicular component of } \mathbf{a} = \left(3\sqrt{2}, 3\sqrt{3}, 3\right) - \left(\frac{\sqrt{2}}{3}, -\frac{\sqrt{3}}{3}, \frac{2}{3}\right)$

Perform the subtraction component-wise:

1. $x$-component: $3\sqrt{2} - \frac{\sqrt{2}}{3} = \frac{9\sqrt{2}}{3} - \frac{\sqrt{2}}{3} = \frac{8\sqrt{2}}{3}$

2. $y$-component: $3\sqrt{3} - \left(-\frac{\sqrt{3}}{3}\right) = \frac{9\sqrt{3}}{3} + \frac{\sqrt{3}}{3} = \frac{10\sqrt{3}}{3}$

3. $z$-component: $3 - \frac{2}{3} = \frac{9}{3} - \frac{2}{3} = \frac{7}{3}$

Thus, the perpendicular component is:

$\left(\frac{8\sqrt{2}}{3}, \frac{10\sqrt{3}}{3}, \frac{7}{3}\right)$

Final Answer:

The correct choice is:

(8/3√2, 10/3√3, 7/3)

Would you like additional clarification or a detailed explanation of any specific step?

Related Questions:

1. What is the geometric meaning of a vector projection?
2. How can you verify that two vectors are orthogonal?
3. What are other ways to calculate vector projections?
4. What is the importance of dot product in determining vector relationships?
5. How can this concept be applied in physics or engineering problems?

Tip: When working with projections, always double-check your scalar and vector calculations for accuracy!