Let `a=i-2j+3k`. If b is a vector such that `a*b=abs(b)^(2)" and "abs(a-b)=sqrt(7)`, then `abs(b)^(2)=`_______

Answer

To solve the problem step by step, we start with the given information:

1. **Given Vector**:
\[ \mathbf{a} = \mathbf{i} - 2\mathbf{j} + 3\mathbf{k} \]

2. **Conditions**:
- \(\mathbf{a} \cdot \mathbf{b} = |\mathbf{b}|^2\)
- \(|\mathbf{a} - \mathbf{b}| = \sqrt{7}\)

### Step 1: Calculate \(|\mathbf{a}|^2\)

First, we find the magnitude squared of vector \(\mathbf{a}\):
\[ |\mathbf{a}|^2 = 1^2 + (-2)^2 + 3^2 = 1 + 4 + 9 = 14 \]

**Hint**: To find the magnitude squared of a vector, square each of its components and sum them up.

### Step 2: Use the second condition

From the second condition, we have:
\[ |\mathbf{a} - \mathbf{b}|^2 = 7 \]

Expanding this using the dot product:
\[ (\mathbf{a} - \mathbf{b}) \cdot (\mathbf{a} - \mathbf{b}) = \mathbf{a} \cdot \mathbf{a} - 2\mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{b} \]
This gives us:
\[ |\mathbf{a}|^2 - 2\mathbf{a} \cdot \mathbf{b} + |\mathbf{b}|^2 = 7 \]

**Hint**: Remember that the dot product of a vector with itself gives its magnitude squared.

### Step 3: Substitute the known values

Substituting \(|\mathbf{a}|^2 = 14\):
\[ 14 - 2\mathbf{a} \cdot \mathbf{b} + |\mathbf{b}|^2 = 7 \]

Rearranging gives:
\[ 14 - 7 = 2\mathbf{a} \cdot \mathbf{b} - |\mathbf{b}|^2 \]
\[ 7 = 2\mathbf{a} \cdot \mathbf{b} - |\mathbf{b}|^2 \]

**Hint**: Rearranging equations can help isolate the terms you want to solve for.

### Step 4: Use the first condition

From the first condition, we know:
\[ \mathbf{a} \cdot \mathbf{b} = |\mathbf{b}|^2 \]
Let \(x = |\mathbf{b}|^2\). Then we can substitute:
\[ 7 = 2x - x \]
This simplifies to:
\[ 7 = x \]

**Hint**: If you let a variable represent a quantity, it can simplify your calculations.

### Step 5: Conclusion

Thus, we find:
\[ |\mathbf{b}|^2 = 7 \]

**Final Answer**:
\[ \boxed{7} \]