For non-zero vectors `veca, vecb and vecc , |(veca xx vecb) .vecc = |veca||vecb||vecc|` holds if and only if

Options

A. `veca.vecb=0 , vecb .vecc=0`
B. `vecb.vecc = 0, vecc, veca =0`
C. `vecc.veca =0, veca,vecb =0`
D. `veca.vecb= vecb.vecc= vecc.veca =0`

Explanation

To solve the problem, we need to analyze the condition under which the equation \( |(\vec{a} \times \vec{b}) \cdot \vec{c}| = |\vec{a}||\vec{b}||\vec{c}| \) holds true for non-zero vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\).

### Step-by-Step Solution:

1. **Understanding the Cross Product**:
The cross product \(\vec{a} \times \vec{b}\) results in a vector that is perpendicular to both \(\vec{a}\) and \(\vec{b}\). The magnitude of this vector is given by:
\[ |\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin(\theta) \]
where \(\theta\) is the angle between \(\vec{a}\) and \(\vec{b}\).

**Hint**: Recall that the magnitude of the cross product depends on the sine of the angle between the two vectors.

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

**Hint**: Remember that the dot product involves the cosine of the angle between the two vectors.

3. **Substituting Magnitudes**:
Substituting the magnitude of the cross product into the dot product expression gives:
\[ |(\vec{a} \times \vec{b}) \cdot \vec{c}| = ||\vec{a}||\vec{b}||\sin(\theta)|||\vec{c}||\cos(\alpha)| \]

**Hint**: This step combines the results from the cross product and the dot product.

4. **Setting Up the Equation**:
Now, we set the absolute value of the dot product equal to the product of the magnitudes:
\[ ||\vec{a}|| |\vec{b}||\vec{c}||\sin(\theta)||\cos(\alpha)| = |\vec{a}||\vec{b}||\vec{c}| \]

**Hint**: This equation shows that the left-hand side must equal the right-hand side.

5. **Simplifying the Equation**:
We can simplify this equation by canceling out the common terms:
\[ |\sin(\theta)||\cos(\alpha)| = 1 \]

**Hint**: This simplification leads to a condition involving sine and cosine.

6. **Analyzing the Conditions**:
The equation \( |\sin(\theta)||\cos(\alpha)| = 1 \) can only hold if:
- \(|\sin(\theta)| = 1\) (which means \(\theta = \frac{\pi}{2}\), i.e., \(\vec{a}\) and \(\vec{b}\) are perpendicular)
- \(|\cos(\alpha)| = 1\) (which means \(\alpha = 0\), i.e., \(\vec{c}\) is parallel to \((\vec{a} \times \vec{b})\))

**Hint**: Consider the geometric implications of sine and cosine being equal to 1.

7. **Conclusion**:
Therefore, the condition holds if and only if:
- \(\vec{a} \perp \vec{b}\) (i.e., \(\vec{a} \cdot \vec{b} = 0\))
- \(\vec{c}\) is perpendicular to both \(\vec{a}\) and \(\vec{b}\) (i.e., \(\vec{c} \cdot \vec{a} = 0\) and \(\vec{c} \cdot \vec{b} = 0\))

Thus, the final answer is:
\[ \vec{a} \cdot \vec{b} = 0, \quad \vec{c} \cdot \vec{a} = 0, \quad \vec{c} \cdot \vec{b} = 0 \]