Let's break down the problem step by step and solve it.

We are given:

$\mathbf{A} = (3.00\hat{i} - 5.00\hat{j}) \, \text{u}, \quad \mathbf{B} = (-1.00\hat{i} + 6.00\hat{j}) \, \text{u}, \quad \mathbf{C} = (-8.00\hat{i} + 9.00\hat{j}) \, \text{u}$

Part (a):

We are asked to find the values of constants $a$ and $b$ that satisfy the vector equation:

$a\mathbf{A} + b\mathbf{B} + \mathbf{C} = 0$

Substitute the vectors for $\mathbf{A}$, $\mathbf{B}$, and $\mathbf{C}$ into this equation:

$a(3.00\hat{i} - 5.00\hat{j}) + b(-1.00\hat{i} + 6.00\hat{j}) + (-8.00\hat{i} + 9.00\hat{j}) = 0$

Now, separate this into the $\hat{i}$ and $\hat{j}$ components:

$\hat{i}$ component:

$3.00a - 1.00b - 8.00 = 0$

$\hat{j}$ component:

$-5.00a + 6.00b + 9.00 = 0$

These are two linear equations with two unknowns, $a$ and $b$. Now we solve these equations.

1. From the $\hat{i}$-component equation: $3.00a - 1.00b = 8.00 \quad \text{(Equation 1)}$

2. From the $\hat{j}$-component equation: $-5.00a + 6.00b = -9.00 \quad \text{(Equation 2)}$

Solving the system of equations:

From Equation 1: $b = 3.00a - 8.00$

Substitute this expression for $b$ into Equation 2: $-5.00a + 6.00(3.00a - 8.00) = -9.00$

Simplify: $-5.00a + 18.00a - 48.00 = -9.00$ $13.00a = 39.00$ $a = 3.00$

Substitute $a = 3.00$ back into the expression for $b$: $b = 3.00(3.00) - 8.00 = 9.00 - 8.00 = 1.00$

Thus, the values of $a$ and $b$ are: $a = 3.00, \quad b = 1.00$

Part (b):

The problem states that only one equation was given, but we still found the values of both $a$ and $b$. This is possible because the equation was separated into two components: the $\hat{i}$-component and the $\hat{j}$-component. These components effectively provided two equations, which are sufficient to solve for the two unknowns, $a$ and $b$. Therefore, the system is well-defined and allows us to find a unique solution.

Do you want more details on any step or have any questions?

Here are some related questions:

1. How do you solve a system of linear equations algebraically?
2. What does it mean when we separate vectors into components?
3. Can we use matrices to solve systems of equations like this one?
4. What does it mean geometrically when two vectors are added to result in zero?
5. How do you verify if your solution to a system of equations is correct?

Tip: Always double-check the number of equations and unknowns in any system of equations to ensure it's solvable.