For three non-coplanar vectors `bara,barb,barc` the relation `|(bara times barb).barc|=|bara||barb||barc|` holds true , if

Options

A. `barb.barc=barc.bara=0`
B. `bara.barb=barb.barc=0`
C. `bara.barb=barb.barc=barc.bara=0`
D. `barc.bara=bara.barb=0`

Answer

To solve the problem, we need to analyze the given relation involving three non-coplanar vectors \(\mathbf{a}\), \(\mathbf{b}\), and \(\mathbf{c}\):

\[ |\mathbf{a} \times \mathbf{b} \cdot \mathbf{c}| = |\mathbf{a}||\mathbf{b}||\mathbf{c}| \]

### Step-by-Step Solution:

1. **Understanding the Left Side**:
The left side of the equation is the absolute value of the dot product of \(\mathbf{c}\) with the cross product \(\mathbf{a} \times \mathbf{b}\). The magnitude of the cross product \(\mathbf{a} \times \mathbf{b}\) is given by:
\[ |\mathbf{a} \times \mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin \theta \]
where \(\theta\) is the angle between vectors \(\mathbf{a}\) and \(\mathbf{b}\).

2. **Dot Product with \(\mathbf{c}\)**:
The dot product of \(\mathbf{c}\) with \(\mathbf{a} \times \mathbf{b}\) can be expressed as:
\[ |\mathbf{a} \times \mathbf{b} \cdot \mathbf{c}| = |\mathbf{a} \times \mathbf{b}||\mathbf{c}|\cos \phi \]
where \(\phi\) is the angle between the vector \(\mathbf{c}\) and the vector \(\mathbf{a} \times \mathbf{b}\).

3. **Equating Both Sides**:
Now we can equate both expressions:
\[ |\mathbf{a}||\mathbf{b}|\sin \theta |\mathbf{c}|\cos \phi = |\mathbf{a}||\mathbf{b}||\mathbf{c}| \]

4. **Simplifying the Equation**:
Dividing both sides by \(|\mathbf{a}||\mathbf{b}||\mathbf{c}|\) (assuming none of them are zero), we get:
\[ \sin \theta \cos \phi = 1 \]

5. **Analyzing the Result**:
The equation \(\sin \theta \cos \phi = 1\) can only hold true if:
- \(\sin \theta = 1\) (which means \(\theta = 90^\circ\))
- \(\cos \phi = 1\) (which means \(\phi = 0^\circ\))

Since \(\theta = 90^\circ\), vectors \(\mathbf{a}\) and \(\mathbf{b}\) are perpendicular. Since \(\phi = 0^\circ\), vector \(\mathbf{c}\) must be parallel to the vector \(\mathbf{a} \times \mathbf{b}\).

6. **Conclusion**:
Therefore, for the relation to hold true, we conclude that:
\[ \mathbf{a} \cdot \mathbf{b} = 0 \quad \text{and} \quad \mathbf{b} \cdot \mathbf{c} = 0 \]
This means that \(\mathbf{b}\) is perpendicular to both \(\mathbf{a}\) and \(\mathbf{c}\).

### Final Answer:
The relation holds true if \(\mathbf{a} \cdot \mathbf{b} = 0\) and \(\mathbf{b} \cdot \mathbf{c} = 0\).

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