If ` vec a =hati +3hatj , vecb = 2hati -hatj - hatk and vec c =mhati +7 hatj +3hatk ` are coplanar then find the value of m.

Answer

To find the value of \( m \) such that the vectors \( \vec{a} = \hat{i} + 3\hat{j} \), \( \vec{b} = 2\hat{i} - \hat{j} - \hat{k} \), and \( \vec{c} = m\hat{i} + 7\hat{j} + 3\hat{k} \) are coplanar, we can use the condition that the scalar triple product of the vectors must be zero. This leads us to the determinant of the matrix formed by the coefficients of the vectors.

### Step-by-step solution:

1. **Write the vectors in component form:**
\[ \vec{a} = \begin{pmatrix} 1 \\ 3 \\ 0 \end{pmatrix}, \quad \vec{b} = \begin{pmatrix} 2 \\ -1 \\ -1 \end{pmatrix}, \quad \vec{c} = \begin{pmatrix} m \\ 7 \\ 3 \end{pmatrix} \]

2. **Set up the determinant for the scalar triple product:**
The vectors are coplanar if the determinant of the matrix formed by these vectors is zero:
\[ \begin{vmatrix} 1 & 3 & 0 \\ 2 & -1 & -1 \\ m & 7 & 3 \end{vmatrix} = 0 \]

3. **Calculate the determinant:**
Using the determinant formula for a \( 3 \times 3 \) matrix:
\[ \text{Det} = 1 \cdot \begin{vmatrix} -1 & -1 \\ 7 & 3 \end{vmatrix} - 3 \cdot \begin{vmatrix} 2 & -1 \\ m & 3 \end{vmatrix} + 0 \cdot \begin{vmatrix} 2 & -1 \\ m & 7 \end{vmatrix} \]

- Calculate the first \( 2 \times 2 \) determinant:
\[ \begin{vmatrix} -1 & -1 \\ 7 & 3 \end{vmatrix} = (-1)(3) - (-1)(7) = -3 + 7 = 4 \]

- Calculate the second \( 2 \times 2 \) determinant:
\[ \begin{vmatrix} 2 & -1 \\ m & 3 \end{vmatrix} = (2)(3) - (-1)(m) = 6 + m = 6 + m \]

- Substitute back into the determinant:
\[ \text{Det} = 1 \cdot 4 - 3 \cdot (6 + m) = 4 - 18 - 3m = -14 - 3m \]

4. **Set the determinant to zero:**
\[ -14 - 3m = 0 \]

5. **Solve for \( m \):**
\[ -3m = 14 \implies m = -\frac{14}{3} \]

Thus, the value of \( m \) is \( -\frac{14}{3} \).