The value of k for which the points A (1, 0, 3), B(-1, 3, 4), C(1, 2, 1) and D(k, 2, 5) are coplanar, are

Options

A. 1
B. 2
C. 0
D. -1

Answer

To find the value of \( k \) for which the points \( A(1, 0, 3) \), \( B(-1, 3, 4) \), \( C(1, 2, 1) \), and \( D(k, 2, 5) \) are coplanar, we will use the concept of the scalar triple product. The points are coplanar if the scalar triple product of the vectors formed by these points is zero.

### Step-by-Step Solution:

1. **Define the Vectors**:
We will first define the vectors \( \vec{AB} \), \( \vec{AC} \), and \( \vec{AD} \).

\[ \vec{AB} = B - A = (-1 - 1, 3 - 0, 4 - 3) = (-2, 3, 1) \]
\[ \vec{AC} = C - A = (1 - 1, 2 - 0, 1 - 3) = (0, 2, -2) \]
\[ \vec{AD} = D - A = (k - 1, 2 - 0, 5 - 3) = (k - 1, 2, 2) \]

2. **Set Up the Scalar Triple Product**:
The scalar triple product \( \vec{AB} \cdot (\vec{AC} \times \vec{AD}) \) must equal zero for the points to be coplanar.

The scalar triple product can be represented as the determinant of the matrix formed by the vectors:
\[ \begin{vmatrix} -2 & 3 & 1 \\ 0 & 2 & -2 \\ k-1 & 2 & 2 \end{vmatrix} = 0 \]

3. **Calculate the Determinant**:
We will calculate the determinant step by step:
\[ = -2 \begin{vmatrix} 2 & -2 \\ 2 & 2 \end{vmatrix} - 3 \begin{vmatrix} 0 & -2 \\ k-1 & 2 \end{vmatrix} + 1 \begin{vmatrix} 0 & 2 \\ k-1 & 2 \end{vmatrix} \]

- Calculate \( \begin{vmatrix} 2 & -2 \\ 2 & 2 \end{vmatrix} = (2)(2) - (-2)(2) = 4 + 4 = 8 \)
- Calculate \( \begin{vmatrix} 0 & -2 \\ k-1 & 2 \end{vmatrix} = (0)(2) - (-2)(k-1) = 2(k-1) \)
- Calculate \( \begin{vmatrix} 0 & 2 \\ k-1 & 2 \end{vmatrix} = (0)(2) - (2)(k-1) = -2(k-1) \)

Now substituting back into the determinant:
\[ = -2(8) - 3(2(k-1)) + 1(-2(k-1)) \]
\[ = -16 - 6(k-1) - 2(k-1) \]
\[ = -16 - 6k + 6 - 2k + 2 \]
\[ = -16 + 8 - 8k = -8 - 8k \]

4. **Set the Determinant to Zero**:
For coplanarity, we set the determinant equal to zero:
\[ -8 - 8k = 0 \]
\[ -8k = 8 \]
\[ k = -1 \]

### Final Answer:
The value of \( k \) for which the points are coplanar is \( k = -1 \).