Let a,b and c be distinct non-negative numbers and the vectors `ahati+ahatj+chatk,hati+hatk,chati+chatj+bhatk` lie in a plane, then the quadratic equation `ax^(2)+2cx+b=0` has

Options

A. real annd equal roots
B. real and unequal roots
C. unreal roots
D. both roots real and positive

Explanation

To solve the problem, we need to analyze the given vectors and their coplanarity condition. Let's break down the solution step by step.

### Step 1: Identify the Vectors
We are given three vectors:
1. **Vector A**: \( \mathbf{A} = a \hat{i} + a \hat{j} + c \hat{k} \)
2. **Vector B**: \( \mathbf{B} = \hat{i} + \hat{k} \)
3. **Vector C**: \( \mathbf{C} = c \hat{i} + c \hat{j} + b \hat{k} \)

### Step 2: Condition for Coplanarity
The three vectors are coplanar if the scalar triple product is zero. This can be expressed using a determinant:
\[ \begin{vmatrix} a & a & c \\ 1 & 0 & 1 \\ c & c & b \end{vmatrix} = 0 \]

### Step 3: Calculate the Determinant
We will compute the determinant:
\[ \begin{vmatrix} a & a & c \\ 1 & 0 & 1 \\ c & c & b \end{vmatrix} \]

Expanding this determinant, we have:
\[ = a \begin{vmatrix} 0 & 1 \\ c & b \end{vmatrix} - a \begin{vmatrix} 1 & 1 \\ c & b \end{vmatrix} + c \begin{vmatrix} 1 & 0 \\ c & c \end{vmatrix} \]

Calculating each of these 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. \( \begin{vmatrix} 1 & 0 \\ c & c \end{vmatrix} = 1 \cdot c - 0 \cdot c = c \)

Putting it all together:
\[ = a(-c) - a(b - c) + c^2 \]
\[ = -ac - ab + ac + c^2 \]
\[ = c^2 - ab \]

### Step 4: Set the Determinant to Zero
Since the vectors are coplanar:
\[ c^2 - ab = 0 \]
This implies:
\[ c^2 = ab \]

### Step 5: Analyze the Quadratic Equation
We are given the quadratic equation:
\[ ax^2 + 2cx + b = 0 \]
To find the nature of the roots, we calculate the discriminant \(D\):
\[ D = (2c)^2 - 4ab \]
Substituting \(c^2 = ab\):
\[ D = 4c^2 - 4ab = 4c^2 - 4c^2 = 0 \]

### Step 6: Conclusion
Since the discriminant \(D = 0\), this means that the quadratic equation has real and equal roots.

### Final Answer
The quadratic equation \(ax^2 + 2cx + b = 0\) has real and equal roots.

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