What is the value of m, if the vector `2 hat(i) - hat(j) + hat(k), hat(i) + 2hat(j)- 3hat(k) and 3hat(i) + m hat(j)+ 5hat(k)` are coplanar?

Options

A. `-2`
B. 2
C. `-4`
D. 4

Answer

To find the value of \( m \) such that the vectors \( \mathbf{a} = 2\hat{i} - \hat{j} + \hat{k} \), \( \mathbf{b} = \hat{i} + 2\hat{j} - 3\hat{k} \), and \( \mathbf{c} = 3\hat{i} + m\hat{j} + 5\hat{k} \) are coplanar, we can use the condition that the scalar triple product of the vectors must be zero.

### Step-by-Step Solution:

1. **Write the vectors in component form**:
\[ \mathbf{a} = \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} 1 \\ 2 \\ -3 \end{pmatrix}, \quad \mathbf{c} = \begin{pmatrix} 3 \\ m \\ 5 \end{pmatrix} \]

2. **Set up the determinant for the scalar triple product**:
The vectors are coplanar if:
\[ \begin{vmatrix} 2 & -1 & 1 \\ 1 & 2 & -3 \\ 3 & m & 5 \end{vmatrix} = 0 \]

3. **Calculate the determinant**:
Expanding the determinant using the first row:
\[ = 2 \begin{vmatrix} 2 & -3 \\ m & 5 \end{vmatrix} - (-1) \begin{vmatrix} 1 & -3 \\ 3 & 5 \end{vmatrix} + 1 \begin{vmatrix} 1 & 2 \\ 3 & m \end{vmatrix} \]

Now calculate each of the 2x2 determinants:
- For \( \begin{vmatrix} 2 & -3 \\ m & 5 \end{vmatrix} = 2 \cdot 5 - (-3) \cdot m = 10 + 3m \)
- For \( \begin{vmatrix} 1 & -3 \\ 3 & 5 \end{vmatrix} = 1 \cdot 5 - (-3) \cdot 3 = 5 + 9 = 14 \)
- For \( \begin{vmatrix} 1 & 2 \\ 3 & m \end{vmatrix} = 1 \cdot m - 2 \cdot 3 = m - 6 \)

Substitute these back into the determinant:
\[ = 2(10 + 3m) + 14 + (m - 6) \]
\[ = 20 + 6m + 14 + m - 6 \]
\[ = 20 + 14 - 6 + 7m = 28 + 7m \]

4. **Set the determinant equal to zero**:
\[ 28 + 7m = 0 \]

5. **Solve for \( m \)**:
\[ 7m = -28 \implies m = -4 \]

### Final Answer:
The value of \( m \) is \( -4 \).