Let `vec(A) = 2hat(i) + hat(k), vec(B) = hat(i) + hat(j) + hat(k)` and `vec(C) = 4hat(i) - 3hat(j) + 7hat(k)`. Determine a vector `vec(R)` satisfying `vec(R) xx vec(B) = vec(C) xx vec(B)` and `vec(R).vec(A) = 0`

Answer

To solve the problem, we need to find a vector \(\vec{R}\) that satisfies two conditions:

1. \(\vec{R} \times \vec{B} = \vec{C} \times \vec{B}\)
2. \(\vec{R} \cdot \vec{A} = 0\)

Given:
\[ \vec{A} = 2\hat{i} + \hat{k}, \quad \vec{B} = \hat{i} + \hat{j} + \hat{k}, \quad \vec{C} = 4\hat{i} - 3\hat{j} + 7\hat{k} \]

### Step 1: Assume the form of \(\vec{R}\)

Let:
\[ \vec{R} = x\hat{i} + y\hat{j} + z\hat{k} \]

### Step 2: Calculate \(\vec{R} \times \vec{B}\)

Using the determinant method for the cross product:
\[ \vec{R} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ x & y & z \\ 1 & 1 & 1 \end{vmatrix} \]

Calculating the determinant:
\[ \vec{R} \times \vec{B} = \hat{i}(y \cdot 1 - z \cdot 1) - \hat{j}(x \cdot 1 - z \cdot 1) + \hat{k}(x \cdot 1 - y \cdot 1) \]
\[ = (y - z)\hat{i} - (x - z)\hat{j} + (x - y)\hat{k} \]

### Step 3: Calculate \(\vec{C} \times \vec{B}\)

Now calculate \(\vec{C} \times \vec{B}\):
\[ \vec{C} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -3 & 7 \\ 1 & 1 & 1 \end{vmatrix} \]

Calculating the determinant:
\[ \vec{C} \times \vec{B} = \hat{i}((-3) \cdot 1 - 7 \cdot 1) - \hat{j}(4 \cdot 1 - 7 \cdot 1) + \hat{k}(4 \cdot 1 - (-3) \cdot 1) \]
\[ = (-3 - 7)\hat{i} - (4 - 7)\hat{j} + (4 + 3)\hat{k} \]
\[ = -10\hat{i} + 3\hat{j} + 7\hat{k} \]

### Step 4: Set the two cross products equal

From the conditions, we have:
\[ (y - z)\hat{i} - (x - z)\hat{j} + (x - y)\hat{k} = -10\hat{i} + 3\hat{j} + 7\hat{k} \]

This gives us the following equations:
1. \(y - z = -10\) (1)
2. \(- (x - z) = 3\) or \(x - z = -3\) (2)
3. \(x - y = 7\) (3)

### Step 5: Solve the system of equations

From equation (1):
\[ y = z - 10 \]

Substituting \(y\) in equation (3):
\[ x - (z - 10) = 7 \implies x - z + 10 = 7 \implies x - z = -3 \quad \text{(which is consistent with equation (2))} \]

Now substituting \(x\) in terms of \(z\):
From equation (2):
\[ x = z - 3 \]

### Step 6: Substitute \(x\) into the equation for \(y\)

Substituting \(x\) into \(y\):
\[ y = (z - 3) - 10 = z - 13 \]

### Step 7: Use the second condition \(\vec{R} \cdot \vec{A} = 0\)

Now we need to satisfy \(\vec{R} \cdot \vec{A} = 0\):
\[ (x\hat{i} + y\hat{j} + z\hat{k}) \cdot (2\hat{i} + \hat{k}) = 0 \]
This gives:
\[ 2x + z = 0 \implies z = -2x \]

### Step 8: Substitute \(z\) back into the equations

Substituting \(z = -2x\) into \(y = z - 10\):
\[ y = -2x - 10 \]

Now substituting \(z = -2x\) into \(x = z - 3\):
\[ x = -2x - 3 \implies 3x = -3 \implies x = -1 \]

### Step 9: Find \(y\) and \(z\)

Now substituting \(x = -1\):
\[ z = -2(-1) = 2 \]
\[ y = -2(-1) - 10 = 2 - 10 = -8 \]

### Step 10: Write the final vector \(\vec{R}\)

Thus, we have:
\[ \vec{R} = -1\hat{i} - 8\hat{j} + 2\hat{k} \]

### Final Answer:
\[ \vec{R} = -\hat{i} - 8\hat{j} + 2\hat{k} \]