A vector of magnitude `sqrt(2)` units and coplanar with vector `3i-j-k and i+j-2k` and perpendicular to vector `2i+2j+k`.....

Answer

To solve the problem, we need to find a vector **V** that satisfies the following conditions:
1. The magnitude of **V** is \( \sqrt{2} \) units.
2. **V** is coplanar with the vectors \( \mathbf{A} = 3\mathbf{i} - \mathbf{j} - \mathbf{k} \) and \( \mathbf{B} = \mathbf{i} + \mathbf{j} - 2\mathbf{k} \).
3. **V** is perpendicular to the vector \( \mathbf{C} = 2\mathbf{i} + 2\mathbf{j} + \mathbf{k} \).

### Step 1: Find a vector in the plane formed by vectors A and B

To find a vector that is coplanar with vectors **A** and **B**, we can use the cross product of **A** and **B** to find a normal vector to the plane formed by them. However, since we need a vector in the plane, we can express **V** as a linear combination of **A** and **B**:

\[ \mathbf{V} = \lambda \mathbf{A} + \mu \mathbf{B} \]

where \( \lambda \) and \( \mu \) are scalars.

### Step 2: Set up the equation for the magnitude of V

The magnitude of vector **V** should equal \( \sqrt{2} \):

\[ |\mathbf{V}| = \sqrt{2} \]

Calculating the magnitude:

\[ |\mathbf{V}| = |\lambda \mathbf{A} + \mu \mathbf{B}| \]

Using the formula for the magnitude of a vector, we have:

\[ |\mathbf{V}| = \sqrt{(\lambda (3) + \mu (1))^2 + (\lambda (-1) + \mu (1))^2 + (\lambda (-1) + \mu (-2))^2} \]

### Step 3: Set up the perpendicularity condition

Since **V** is perpendicular to **C**, we have:

\[ \mathbf{V} \cdot \mathbf{C} = 0 \]

Calculating the dot product:

\[ (\lambda \mathbf{A} + \mu \mathbf{B}) \cdot (2\mathbf{i} + 2\mathbf{j} + \mathbf{k}) = 0 \]

This gives us another equation to work with.

### Step 4: Solve the system of equations

Now we have two equations:
1. The magnitude equation.
2. The perpendicularity equation.

We can solve these equations simultaneously to find the values of \( \lambda \) and \( \mu \).

### Step 5: Substitute values and simplify

After substituting values and simplifying, we can find the specific values of \( \lambda \) and \( \mu \) that satisfy both conditions.

### Step 6: Construct the vector V

Finally, we can construct the vector **V** using the values of \( \lambda \) and \( \mu \) we found.