If `vecA=(1,1,1) and vecC=(0,1,-1)` are given vectors then find a vector `vecB` satisfying equations `vecAxxvecB=vecC and vecA.vecB=3`

Explanation

To find the vector \(\vec{B}\) that satisfies the equations \(\vec{A} \times \vec{B} = \vec{C}\) and \(\vec{A} \cdot \vec{B} = 3\), we will follow these steps:

### Step 1: Define the vectors
Given:
\[ \vec{A} = (1, 1, 1) \quad \text{and} \quad \vec{C} = (0, 1, -1) \]
Let the vector \(\vec{B} = (x, y, z)\).

### Step 2: Use the dot product equation
From the equation \(\vec{A} \cdot \vec{B} = 3\):
\[ \vec{A} \cdot \vec{B} = 1 \cdot x + 1 \cdot y + 1 \cdot z = x + y + z = 3 \]
This gives us our first equation:
\[ (1) \quad x + y + z = 3 \]

### Step 3: Use the cross product equation
Now, we will compute the cross product \(\vec{A} \times \vec{B}\):
\[ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ x & y & z \end{vmatrix} \]
Calculating the determinant:
\[ \vec{A} \times \vec{B} = \hat{i}(1 \cdot z - 1 \cdot y) - \hat{j}(1 \cdot z - 1 \cdot x) + \hat{k}(1 \cdot y - 1 \cdot x) \]
This simplifies to:
\[ \vec{A} \times \vec{B} = (z - y) \hat{i} + (x - z) \hat{j} + (y - x) \hat{k} \]
Setting this equal to \(\vec{C} = (0, 1, -1)\), we have:
\[ (2) \quad z - y = 0 \\ (3) \quad x - z = 1 \\ (4) \quad y - x = -1 \]

### Step 4: Solve the system of equations
From equation (2):
\[ z = y \]
Substituting \(z = y\) into equation (3):
\[ x - y = 1 \quad \Rightarrow \quad x = y + 1 \]
Substituting \(x = y + 1\) into equation (4):
\[ y - (y + 1) = -1 \quad \Rightarrow \quad -1 = -1 \quad \text{(which is always true)} \]
Now substitute \(x = y + 1\) and \(z = y\) into equation (1):
\[ (y + 1) + y + y = 3 \\ 3y + 1 = 3 \\ 3y = 2 \\ y = \frac{2}{3} \]
Now, substituting back to find \(x\) and \(z\):
\[ z = y = \frac{2}{3} \\ x = y + 1 = \frac{2}{3} + 1 = \frac{5}{3} \]

### Step 5: Write the final vector \(\vec{B}\)
Thus, the vector \(\vec{B}\) is:
\[ \vec{B} = \left(\frac{5}{3}, \frac{2}{3}, \frac{2}{3}\right) \]

### Summary
The vector \(\vec{B}\) that satisfies the given conditions is:
\[ \vec{B} = \left(\frac{5}{3}, \frac{2}{3}, \frac{2}{3}\right) \]