Show that the four points A,B,C and D with position vectors `4hati+5hatj+hatk, -hatj-hatk, 3hati+9hatj+4hatk and 4(-hati+hatj+hatk)` respectively, are coplanar.

Answer

To show that the four points A, B, C, and D with position vectors \( \mathbf{A} = 4\hat{i} + 5\hat{j} + \hat{k} \), \( \mathbf{B} = -\hat{j} - \hat{k} \), \( \mathbf{C} = 3\hat{i} + 9\hat{j} + 4\hat{k} \), and \( \mathbf{D} = 4(-\hat{i} + \hat{j} + \hat{k}) \) are coplanar, we can use the concept of the scalar triple product. The points are coplanar if the volume of the parallelepiped formed by the vectors \( \overrightarrow{AB} \), \( \overrightarrow{AC} \), and \( \overrightarrow{AD} \) is zero.

### Step 1: Find the vectors \( \overrightarrow{AB} \), \( \overrightarrow{AC} \), and \( \overrightarrow{AD} \).

\[ \overrightarrow{AB} = \mathbf{B} - \mathbf{A} = (-\hat{j} - \hat{k}) - (4\hat{i} + 5\hat{j} + \hat{k}) = -4\hat{i} - 6\hat{j} - 2\hat{k} \]

\[ \overrightarrow{AC} = \mathbf{C} - \mathbf{A} = (3\hat{i} + 9\hat{j} + 4\hat{k}) - (4\hat{i} + 5\hat{j} + \hat{k}) = -\hat{i} + 4\hat{j} + 3\hat{k} \]

\[ \overrightarrow{AD} = \mathbf{D} - \mathbf{A} = 4(-\hat{i} + \hat{j} + \hat{k}) - (4\hat{i} + 5\hat{j} + \hat{k}) = -8\hat{i} - \hat{j} + 3\hat{k} \]

### Step 2: Set up the matrix for the scalar triple product.

The scalar triple product can be calculated using the determinant of the matrix formed by the vectors \( \overrightarrow{AB} \), \( \overrightarrow{AC} \), and \( \overrightarrow{AD} \):

\[ \begin{vmatrix} -4 & -6 & -2 \\ -1 & 4 & 3 \\ -8 & -1 & 3 \end{vmatrix} \]

### Step 3: Calculate the determinant.

Calculating the determinant:

\[ = -4 \begin{vmatrix} 4 & 3 \\ -1 & 3 \end{vmatrix} + 6 \begin{vmatrix} -1 & 3 \\ -8 & 3 \end{vmatrix} - 2 \begin{vmatrix} -1 & 4 \\ -8 & -1 \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} 4 & 3 \\ -1 & 3 \end{vmatrix} = (4)(3) - (3)(-1) = 12 + 3 = 15 \)

2. \( \begin{vmatrix} -1 & 3 \\ -8 & 3 \end{vmatrix} = (-1)(3) - (3)(-8) = -3 + 24 = 21 \)

3. \( \begin{vmatrix} -1 & 4 \\ -8 & -1 \end{vmatrix} = (-1)(-1) - (4)(-8) = 1 + 32 = 33 \)

Substituting back into the determinant:

\[ = -4(15) + 6(21) - 2(33) \]
\[ = -60 + 126 - 66 \]
\[ = 0 \]

### Conclusion:

Since the determinant is zero, the volume of the parallelepiped is zero, which means that the points A, B, C, and D are coplanar.