If `veca=hatj-hatk` and `vecc=hati+hatj+hatk` are given vectors, and `vecb` is such that `veca.vecb=3` and `vecaxxvecb+vecc=0` than `|vecb|^(2)` is equal to ………………

Answer

To solve the given problem step by step, we will follow the outlined approach to find the magnitude squared of vector \(\vec{b}\).

### Given:
- \(\vec{a} = \hat{j} - \hat{k}\)
- \(\vec{c} = \hat{i} + \hat{j} + \hat{k}\)
- \(\vec{a} \cdot \vec{b} = 3\)
- \(\vec{a} \times \vec{b} + \vec{c} = 0\)

### Step 1: Write \(\vec{b}\) in component form
Let \(\vec{b} = b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k}\).

### Step 2: Use the dot product condition
From the condition \(\vec{a} \cdot \vec{b} = 3\):
\[ (\hat{j} - \hat{k}) \cdot (b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k}) = 3 \]
Calculating the dot product:
\[ 0 \cdot b_1 + 1 \cdot b_2 - 1 \cdot b_3 = 3 \]
This simplifies to:
\[ b_2 - b_3 = 3 \quad \text{(Equation 1)} \]

### Step 3: Use the cross product condition
From the condition \(\vec{a} \times \vec{b} + \vec{c} = 0\):
\[ \vec{a} \times \vec{b} = -\vec{c} \]
Calculating \(\vec{a} \times \vec{b}\):
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & -1 \\ b_1 & b_2 & b_3 \end{vmatrix} \]
Calculating the determinant:
\[ \vec{a} \times \vec{b} = \hat{i} \left(1 \cdot b_3 - (-1) \cdot b_2\right) - \hat{j} \left(0 \cdot b_3 - (-1) \cdot b_1\right) + \hat{k} \left(0 \cdot b_2 - 1 \cdot b_1\right) \]
This simplifies to:
\[ \vec{a} \times \vec{b} = (b_3 + b_2) \hat{i} - b_1 \hat{j} - b_1 \hat{k} \]
Setting this equal to \(-\vec{c}\):
\[ (b_3 + b_2) \hat{i} - b_1 \hat{j} - b_1 \hat{k} = -(\hat{i} + \hat{j} + \hat{k}) \]
This gives us the following equations:
1. \(b_3 + b_2 = -1\) \quad \text{(Equation 2)}
2. \(-b_1 = -1 \Rightarrow b_1 = 1\) \quad \text{(Equation 3)}

### Step 4: Solve the equations
Substituting \(b_1 = 1\) into Equation 1:
\[ b_2 - b_3 = 3 \]
Substituting \(b_2 = -1 - b_3\) from Equation 2 into this:
\[ (-1 - b_3) - b_3 = 3 \]
This simplifies to:
\[ -1 - 2b_3 = 3 \Rightarrow -2b_3 = 4 \Rightarrow b_3 = -2 \]
Now substituting \(b_3 = -2\) back into Equation 2:
\[ b_2 + (-2) = -1 \Rightarrow b_2 = 1 \]

### Step 5: Write \(\vec{b}\)
Now we have:
\[ b_1 = 1, \quad b_2 = 1, \quad b_3 = -2 \]
Thus, \(\vec{b} = 1 \hat{i} + 1 \hat{j} - 2 \hat{k}\).

### Step 6: Find \(|\vec{b}|^2\)
Calculating the magnitude squared:
\[ |\vec{b}|^2 = b_1^2 + b_2^2 + b_3^2 = 1^2 + 1^2 + (-2)^2 = 1 + 1 + 4 = 6 \]

### Final Answer:
\(|\vec{b}|^2 = 6\)

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