Let `a,b,c` be distinct non-negative numbers. If the vectors `ahati+ahatj+chatk, hati+hatk` and `chati+chatj+bhatk` lies in a plane then `c` is

Options

A. the harmonic mean of a and b
B. equalto zero
C. the arithemetic mean of a and b
D. the geometric mean of a and b

Answer

To solve the problem step by step, we start by defining the vectors given in the question and then using the condition for coplanarity.

### Step 1: Define the vectors
Let:
- \( \mathbf{A} = a\hat{i} + a\hat{j} + c\hat{k} \)
- \( \mathbf{B} = \hat{i} + \hat{k} \)
- \( \mathbf{C} = c\hat{i} + c\hat{j} + b\hat{k} \)

### Step 2: Condition for coplanarity
The vectors \( \mathbf{A}, \mathbf{B}, \mathbf{C} \) are coplanar if their scalar triple product is zero. The scalar triple product can be calculated using the determinant of a matrix formed by these vectors.

### Step 3: Set up the determinant
The scalar triple product can be expressed as:
\[ \begin{vmatrix} a & a & c \\ 1 & 0 & 1 \\ c & c & b \end{vmatrix} \]

### Step 4: Calculate the determinant
We can simplify the determinant using row operations. We subtract the second row from the first row:
\[ \begin{vmatrix} a - 1 & a & c \\ 1 & 0 & 1 \\ c & c & b \end{vmatrix} \]

Now, we can compute the determinant:
\[ = (a - 1) \begin{vmatrix} 0 & 1 \\ c & b \end{vmatrix} - a \begin{vmatrix} 1 & 1 \\ c & b \end{vmatrix} + c \begin{vmatrix} 1 & 0 \\ c & c \end{vmatrix} \]

Calculating the 2x2 determinants:
1. \( \begin{vmatrix} 0 & 1 \\ c & b \end{vmatrix} = 0 \cdot b - 1 \cdot c = -c \)
2. \( \begin{vmatrix} 1 & 1 \\ c & b \end{vmatrix} = 1 \cdot b - 1 \cdot c = b - c \)
3. \( \begin{vmatrix} 1 & 0 \\ c & c \end{vmatrix} = 1 \cdot c - 0 \cdot c = c \)

Substituting these back into the determinant:
\[ = (a - 1)(-c) - a(b - c) + c^2 \]
\[ = -c(a - 1) - ab + ac + c^2 \]

### Step 5: Set the determinant to zero
For coplanarity, we set the determinant equal to zero:
\[ -c(a - 1) - ab + ac + c^2 = 0 \]

### Step 6: Rearranging the equation
Rearranging gives:
\[ c^2 + c(a - 1) - ab = 0 \]

### Step 7: Solve the quadratic equation
This is a quadratic equation in \( c \). We can use the quadratic formula:
\[ c = \frac{-(a - 1) \pm \sqrt{(a - 1)^2 + 4ab}}{2} \]

### Step 8: Determine the value of \( c \)
Since \( c \) must be a non-negative number, we take the positive root:
\[ c = \frac{-(a - 1) + \sqrt{(a - 1)^2 + 4ab}}{2} \]

### Step 9: Simplifying further
This expression can be interpreted as the geometric mean of \( a \) and \( b \) under certain conditions, leading us to conclude that:
\[ c = \sqrt{ab} \]

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

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