If `veca=hati+hatj+hatk, veca.vecb=1` and `vecaxxvecb=hatj-hatk` then `vecb`

Options

A. `hati-hatj+hatk`
B. `2hatj-hatk`
C. `hati`
D. `2hati`

Explanation

To solve the problem, we need to find the vector \(\vec{b}\) given the conditions:

1. \(\vec{a} = \hat{i} + \hat{j} + \hat{k}\)
2. \(\vec{a} \cdot \vec{b} = 1\)
3. \(\vec{a} \times \vec{b} = \hat{j} - \hat{k}\)

### Step 1: Write down the vectors
We have:
\[ \vec{a} = \hat{i} + \hat{j} + \hat{k} \]
Let \(\vec{b} = x\hat{i} + y\hat{j} + z\hat{k}\), where \(x\), \(y\), and \(z\) are the components of \(\vec{b}\).

### Step 2: Use the dot product condition
From the condition \(\vec{a} \cdot \vec{b} = 1\):
\[ (\hat{i} + \hat{j} + \hat{k}) \cdot (x\hat{i} + y\hat{j} + z\hat{k}) = 1 \]
Calculating the dot product:
\[ x + y + z = 1 \quad \text{(Equation 1)} \]

### Step 3: Use the cross product condition
Now, we calculate \(\vec{a} \times \vec{b}\):
\[ \vec{a} \times \vec{b} = (\hat{i} + \hat{j} + \hat{k}) \times (x\hat{i} + y\hat{j} + z\hat{k}) \]
Using the determinant method for the cross product:
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ x & y & z \end{vmatrix} \]
Calculating the determinant:
\[ = \hat{i} \begin{vmatrix} 1 & 1 \\ y & z \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 1 \\ x & z \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 1 \\ x & y \end{vmatrix} \]
Calculating the minors:
\[ = \hat{i}(1z - 1y) - \hat{j}(1z - 1x) + \hat{k}(1y - 1x) \]
\[ = (z - y)\hat{i} - (z - x)\hat{j} + (y - x)\hat{k} \]
Setting this equal to \(\hat{j} - \hat{k}\):
\[ (z - y)\hat{i} - (z - x)\hat{j} + (y - x)\hat{k} = \hat{j} - \hat{k} \]

### Step 4: Equate coefficients
From the equation above, we can equate coefficients:
1. For \(\hat{i}\): \(z - y = 0 \quad \Rightarrow \quad z = y \quad \text{(Equation 2)}\)
2. For \(\hat{j}\): \(- (z - x) = 1 \quad \Rightarrow \quad z - x = -1 \quad \Rightarrow \quad x = z + 1 \quad \text{(Equation 3)}\)
3. For \(\hat{k}\): \(y - x = -1 \quad \Rightarrow \quad y - x = -1 \quad \Rightarrow \quad x = y + 1 \quad \text{(Equation 4)}\)

### Step 5: Substitute and solve
From Equation 2, we have \(z = y\). Substitute \(z\) in Equation 3:
\[ x = y + 1 \]
Now substitute \(x\) and \(z\) in Equation 1:
\[ (y + 1) + y + y = 1 \]
\[ 3y + 1 = 1 \quad \Rightarrow \quad 3y = 0 \quad \Rightarrow \quad y = 0 \]
Then from \(y = 0\):
\[ z = 0 \quad \text{and} \quad x = 1 \]

### Final Result
Thus, we have:
\[ \vec{b} = 1\hat{i} + 0\hat{j} + 0\hat{k} = \hat{i} \]

### Summary
The vector \(\vec{b}\) is:
\[ \vec{b} = \hat{i} \]