IF a,b,c are three real numbers not all equal and the vectors `barx=ahati+bhatj+chatk,bary=bhati+chatj+ahatk,barz=c hati+ahatj+bhatk` are coplanar then `barx.bary+bary.barz+barz.barx` is necessarily……

Options

A. positive
B. non-negative
C. non-positive
D. negative

Answer

To solve the problem, we need to determine the value of \( \bar{x} \cdot \bar{y} + \bar{y} \cdot \bar{z} + \bar{z} \cdot \bar{x} \) given that the vectors \( \bar{x} \), \( \bar{y} \), and \( \bar{z} \) are coplanar.

### Step-by-Step Solution:

1. **Define the Vectors**:
The vectors are defined as follows:
\[ \bar{x} = a \hat{i} + b \hat{j} + c \hat{k} \]
\[ \bar{y} = b \hat{i} + c \hat{j} + a \hat{k} \]
\[ \bar{z} = c \hat{i} + a \hat{j} + b \hat{k} \]

2. **Calculate the Dot Products**:
We will calculate each dot product separately.

- **Dot Product \( \bar{x} \cdot \bar{y} \)**:
\[ \bar{x} \cdot \bar{y} = (a \hat{i} + b \hat{j} + c \hat{k}) \cdot (b \hat{i} + c \hat{j} + a \hat{k}) = ab + bc + ca \]

- **Dot Product \( \bar{y} \cdot \bar{z} \)**:
\[ \bar{y} \cdot \bar{z} = (b \hat{i} + c \hat{j} + a \hat{k}) \cdot (c \hat{i} + a \hat{j} + b \hat{k}) = bc + ac + ab \]

- **Dot Product \( \bar{z} \cdot \bar{x} \)**:
\[ \bar{z} \cdot \bar{x} = (c \hat{i} + a \hat{j} + b \hat{k}) \cdot (a \hat{i} + b \hat{j} + c \hat{k}) = ac + ab + bc \]

3. **Sum the Dot Products**:
Now we sum the results of the dot products:
\[ \bar{x} \cdot \bar{y} + \bar{y} \cdot \bar{z} + \bar{z} \cdot \bar{x} = (ab + bc + ca) + (bc + ac + ab) + (ac + ab + bc) \]
This simplifies to:
\[ = 3(ab + bc + ca) \]

4. **Condition for Coplanarity**:
Since the vectors \( \bar{x}, \bar{y}, \bar{z} \) are coplanar, their scalar triple product must be zero:
\[ \bar{x} \cdot (\bar{y} \times \bar{z}) = 0 \]
This implies that:
\[ ab + bc + ca = 0 \]

5. **Final Result**:
Substituting \( ab + bc + ca = 0 \) into our earlier result:
\[ 3(ab + bc + ca) = 3 \cdot 0 = 0 \]

Thus, the value of \( \bar{x} \cdot \bar{y} + \bar{y} \cdot \bar{z} + \bar{z} \cdot \bar{x} \) is necessarily **0**.