If vectors i + j + k,i - j+k and 2i + 3j + mk are coplanar, then m =

Options

A. 2
B. 3
C. `-3`
D. `-2`

Answer

To determine the value of \( m \) for which the vectors \( \mathbf{a} = \mathbf{i} + \mathbf{j} + \mathbf{k} \), \( \mathbf{b} = \mathbf{i} - \mathbf{j} + \mathbf{k} \), and \( \mathbf{c} = 2\mathbf{i} + 3\mathbf{j} + m\mathbf{k} \) are coplanar, we can use the concept of the scalar triple product. The vectors are coplanar if their scalar triple product is zero.

### Step-by-Step Solution:

1. **Define the Vectors**:
\[ \mathbf{a} = \mathbf{i} + \mathbf{j} + \mathbf{k} \]
\[ \mathbf{b} = \mathbf{i} - \mathbf{j} + \mathbf{k} \]
\[ \mathbf{c} = 2\mathbf{i} + 3\mathbf{j} + m\mathbf{k} \]

2. **Set Up the Scalar Triple Product**:
The scalar triple product can be expressed as:
\[ \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = 0 \]
This can be calculated using the determinant of a matrix formed by the coefficients of the vectors.

3. **Construct the Determinant**:
The determinant is given by:
\[ \begin{vmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 2 & 3 & m \end{vmatrix} \]

4. **Calculate the Determinant**:
Expanding the determinant:
\[ = 1 \cdot \begin{vmatrix} -1 & 1 \\ 3 & m \end{vmatrix} - 1 \cdot \begin{vmatrix} 1 & 1 \\ 2 & m \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & -1 \\ 2 & 3 \end{vmatrix} \]

Now calculating each of these 2x2 determinants:
- First determinant:
\[ = (-1)(m) - (1)(3) = -m - 3 \]
- Second determinant:
\[ = (1)(m) - (1)(2) = m - 2 \]
- Third determinant:
\[ = (1)(3) - (-1)(2) = 3 + 2 = 5 \]

5. **Combine the Results**:
Putting it all together:
\[ -m - 3 - (m - 2) + 5 = 0 \]
Simplifying this:
\[ -m - 3 - m + 2 + 5 = 0 \]
\[ -2m + 4 = 0 \]

6. **Solve for \( m \)**:
\[ -2m = -4 \implies m = 2 \]

### Final Answer:
The value of \( m \) is \( 2 \).