Let a,b,c be distinct non zero numbers. If the vectors `a vec i + a vecj +c vec k`, `vec i + vec k` and `c vec i + c vec j + b vec k` lie in a plane then 'c' is

Options

A. the AM of a and b
B. the GM of a and b
C. the HM of a and b
D. equal to zero

Answer

To solve the problem, we need to determine the value of \( c \) given that the vectors \( \vec{A} = a \vec{i} + a \vec{j} + c \vec{k} \), \( \vec{B} = \vec{i} + \vec{k} \), and \( \vec{C} = c \vec{i} + c \vec{j} + b \vec{k} \) are coplanar. The condition for coplanarity of three vectors is that their scalar triple product is zero.

### Step-by-Step Solution:

1. **Write the Vectors**:
\[ \vec{A} = a \vec{i} + a \vec{j} + c \vec{k} \]
\[ \vec{B} = \vec{i} + \vec{k} \]
\[ \vec{C} = c \vec{i} + c \vec{j} + b \vec{k} \]

2. **Set Up the Scalar Triple Product**:
The scalar triple product can be expressed as the determinant of a matrix formed by the coefficients of \( \vec{i} \), \( \vec{j} \), and \( \vec{k} \):
\[ \begin{vmatrix} a & a & c \\ 1 & 0 & 1 \\ c & c & b \end{vmatrix} = 0 \]

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

- Calculate the first determinant:
\[ = a(0 \cdot b - 1 \cdot c) = -ac \]

- Calculate the second determinant:
\[ = -a(1 \cdot b - 1 \cdot c) = -a(b - c) = -ab + ac \]

- Calculate the third determinant:
\[ = c(1 \cdot c - 0 \cdot c) = c^2 \]

4. **Combine the Results**:
Putting it all together:
\[ -ac - ab + ac + c^2 = 0 \]
Simplifying this gives:
\[ -ab + c^2 = 0 \]
Thus, we have:
\[ c^2 = ab \]

5. **Solve for \( c \)**:
Taking the square root of both sides, we find:
\[ c = \sqrt{ab} \]
Since \( a \) and \( b \) are distinct non-zero numbers, we conclude that:
\[ c = \text{Geometric Mean of } a \text{ and } b \]

### Conclusion:
Thus, the value of \( c \) is the geometric mean of \( a \) and \( b \).