Show that the following vectors are coplanar :
`hat(i)-hat(j)+hat(k), 6hat(i)-hat(k)` and `4hat(i)+2hat(j)-3hat(k)`

Answer

To show that the vectors \( \mathbf{A} = \hat{i} - \hat{j} + \hat{k} \), \( \mathbf{B} = 6\hat{i} - \hat{k} \), and \( \mathbf{C} = 4\hat{i} + 2\hat{j} - 3\hat{k} \) are coplanar, we can use the determinant method. The vectors are coplanar if the determinant of the matrix formed by their components is equal to zero.

### Step-by-step Solution:

1. **Identify the Vectors**:
- Let \( \mathbf{A} = \hat{i} - \hat{j} + \hat{k} \)
- Let \( \mathbf{B} = 6\hat{i} - \hat{k} \)
- Let \( \mathbf{C} = 4\hat{i} + 2\hat{j} - 3\hat{k} \)

2. **Write the Coefficients**:
- The coefficients for vector \( \mathbf{A} \) are \( (1, -1, 1) \)
- The coefficients for vector \( \mathbf{B} \) are \( (6, 0, -1) \)
- The coefficients for vector \( \mathbf{C} \) are \( (4, 2, -3) \)

3. **Set Up the Determinant**:
We need to compute the determinant of the following matrix:

\[ \begin{vmatrix} 1 & -1 & 1 \\ 6 & 0 & -1 \\ 4 & 2 & -3 \end{vmatrix} \]

4. **Calculate the Determinant**:
We can expand the determinant using the first row:

\[ \text{Det} = 1 \cdot \begin{vmatrix} 0 & -1 \\ 2 & -3 \end{vmatrix} - (-1) \cdot \begin{vmatrix} 6 & -1 \\ 4 & -3 \end{vmatrix} + 1 \cdot \begin{vmatrix} 6 & 0 \\ 4 & 2 \end{vmatrix} \]

Now calculating each of these 2x2 determinants:

- First determinant:
\[ \begin{vmatrix} 0 & -1 \\ 2 & -3 \end{vmatrix} = (0)(-3) - (-1)(2) = 0 + 2 = 2 \]

- Second determinant:
\[ \begin{vmatrix} 6 & -1 \\ 4 & -3 \end{vmatrix} = (6)(-3) - (-1)(4) = -18 + 4 = -14 \]

- Third determinant:
\[ \begin{vmatrix} 6 & 0 \\ 4 & 2 \end{vmatrix} = (6)(2) - (0)(4) = 12 - 0 = 12 \]

Now substituting back into the determinant calculation:

\[ \text{Det} = 1 \cdot 2 + 1 \cdot 14 + 1 \cdot 12 = 2 + 14 + 12 = 28 \]

5. **Conclusion**:
Since the determinant is not equal to zero, the vectors \( \mathbf{A}, \mathbf{B}, \mathbf{C} \) are **not coplanar**.