If `veca=veci+vecj+veck, veca.vecb=1 and vecaxxvecb=vecj-veck,` then `vecb` equals

Options

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

Answer

To solve the problem, we need to find the vector \( \vec{b} \) given the conditions on the vectors \( \vec{a} \) and \( \vec{b} \).

### Step 1: Define the vectors
Given:
\[ \vec{a} = \vec{i} + \vec{j} + \vec{k} \]
Let:
\[ \vec{b} = x \vec{i} + y \vec{j} + z \vec{k} \]

### Step 2: Use the dot product condition
We know that:
\[ \vec{a} \cdot \vec{b} = 1 \]
Calculating the dot product:
\[ (\vec{i} + \vec{j} + \vec{k}) \cdot (x \vec{i} + y \vec{j} + z \vec{k}) = x + y + z = 1 \]
So we have our first equation:
\[ x + y + z = 1 \quad \text{(1)} \]

### Step 3: Use the cross product condition
We also know:
\[ \vec{a} \times \vec{b} = \vec{j} - \vec{k} \]
Calculating the cross product:
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & 1 & 1 \\ x & y & z \end{vmatrix} \]
Calculating the determinant:
\[ = \vec{i} \begin{vmatrix} 1 & 1 \\ y & z \end{vmatrix} - \vec{j} \begin{vmatrix} 1 & 1 \\ x & z \end{vmatrix} + \vec{k} \begin{vmatrix} 1 & 1 \\ x & y \end{vmatrix} \]
Calculating each of the 2x2 determinants:
\[ = \vec{i} (1 \cdot z - 1 \cdot y) - \vec{j} (1 \cdot z - 1 \cdot x) + \vec{k} (1 \cdot y - 1 \cdot x) \]
This simplifies to:
\[ = (z - y) \vec{i} - (z - x) \vec{j} + (y - x) \vec{k} \]
Setting this equal to \( \vec{j} - \vec{k} \):
\[ (z - y) \vec{i} - (z - x) \vec{j} + (y - x) \vec{k} = \vec{j} - \vec{k} \]

### Step 4: Compare coefficients
From the equation above, we can compare coefficients:
1. Coefficient of \( \vec{i} \): \( z - y = 0 \) (2)
2. Coefficient of \( \vec{j} \): \( -(z - x) = 1 \) or \( z - x = -1 \) (3)
3. Coefficient of \( \vec{k} \): \( y - x = -1 \) (4)

### Step 5: Solve the system of equations
From equation (2):
\[ z = y \quad \text{(5)} \]
Substituting (5) into (3):
\[ y - x = -1 \implies y = x - 1 \quad \text{(6)} \]
Substituting (6) into (1):
\[ x + (x - 1) + y = 1 \implies x + (x - 1) + (x - 1) = 1 \]
This simplifies to:
\[ 3x - 2 = 1 \implies 3x = 3 \implies x = 1 \]
Now substituting \( x = 1 \) into (6):
\[ y = 1 - 1 = 0 \]
And substituting \( y = 0 \) into (5):
\[ z = 0 \]

### Step 6: Write the final vector \( \vec{b} \)
Thus, we have:
\[ \vec{b} = 1 \vec{i} + 0 \vec{j} + 0 \vec{k} = \vec{i} \]

### Final Answer
\[ \vec{b} = \vec{i} \]