If `hatd` is a unit vectors such that `hatd=lamdabarb times barc+mubarc times bara+vbara times barb` then `|(hatd.bara)(barb times barc)+(bard.barb) (barc times bara)+(bard.barc) (bara times barb)|` is equal to

Options

A. `|[bara barb barc]|`
B. 1
C. `3|[bara barb barc]|`
D. None of these

Answer

To solve the problem, we need to evaluate the expression given in the question step by step.

### Given:
\[ \hat{d} = \lambda (\mathbf{b} \times \mathbf{c}) + \mu (\mathbf{c} \times \mathbf{a}) + \nu (\mathbf{a} \times \mathbf{b}) \]
where \(\hat{d}\) is a unit vector.

We need to find:
\[ |\hat{d} \cdot \mathbf{a} (\mathbf{b} \times \mathbf{c}) + \hat{d} \cdot \mathbf{b} (\mathbf{c} \times \mathbf{a}) + \hat{d} \cdot \mathbf{c} (\mathbf{a} \times \mathbf{b})| \]

### Step 1: Calculate \(\hat{d} \cdot \mathbf{a}\)
Using the expression for \(\hat{d}\):
\[ \hat{d} \cdot \mathbf{a} = \lambda (\mathbf{b} \times \mathbf{c}) \cdot \mathbf{a} + \mu (\mathbf{c} \times \mathbf{a}) \cdot \mathbf{a} + \nu (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{a} \]
The second and third terms vanish because the dot product of a vector with itself is zero:
\[ \hat{d} \cdot \mathbf{a} = \lambda (\mathbf{b} \times \mathbf{c}) \cdot \mathbf{a} \]

### Step 2: Calculate \(\hat{d} \cdot \mathbf{b}\)
Similarly,
\[ \hat{d} \cdot \mathbf{b} = \lambda (\mathbf{b} \times \mathbf{c}) \cdot \mathbf{b} + \mu (\mathbf{c} \times \mathbf{a}) \cdot \mathbf{b} + \nu (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{b} \]
Again, the first and third terms vanish:
\[ \hat{d} \cdot \mathbf{b} = \mu (\mathbf{c} \times \mathbf{a}) \cdot \mathbf{b} \]

### Step 3: Calculate \(\hat{d} \cdot \mathbf{c}\)
Now for \(\hat{d} \cdot \mathbf{c}\):
\[ \hat{d} \cdot \mathbf{c} = \lambda (\mathbf{b} \times \mathbf{c}) \cdot \mathbf{c} + \mu (\mathbf{c} \times \mathbf{a}) \cdot \mathbf{c} + \nu (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c} \]
Again, the first and second terms vanish:
\[ \hat{d} \cdot \mathbf{c} = \nu (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c} \]

### Step 4: Substitute back into the expression
Now substitute these results back into the original expression:
\[ |\hat{d} \cdot \mathbf{a} (\mathbf{b} \times \mathbf{c}) + \hat{d} \cdot \mathbf{b} (\mathbf{c} \times \mathbf{a}) + \hat{d} \cdot \mathbf{c} (\mathbf{a} \times \mathbf{b})| \]
This becomes:
\[ |\lambda (\mathbf{b} \times \mathbf{c}) \cdot \mathbf{a} (\mathbf{b} \times \mathbf{c}) + \mu (\mathbf{c} \times \mathbf{a}) \cdot \mathbf{b} (\mathbf{c} \times \mathbf{a}) + \nu (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c} (\mathbf{a} \times \mathbf{b})| \]

### Step 5: Recognize the scalar triple product
Notice that:
\[ (\mathbf{b} \times \mathbf{c}) \cdot \mathbf{a} = \text{Volume of the parallelepiped formed by } \mathbf{a}, \mathbf{b}, \mathbf{c} = [\mathbf{a}, \mathbf{b}, \mathbf{c}] \]
Thus, we can express the entire expression as:
\[ |[\mathbf{a}, \mathbf{b}, \mathbf{c}]| \]

### Step 6: Final Result
Since \(\hat{d}\) is a unit vector, we have:
\[ |[\mathbf{a}, \mathbf{b}, \mathbf{c}]| = 1 \]

### Conclusion
The final answer is:
\[ |[\mathbf{a}, \mathbf{b}, \mathbf{c}]| = 1 \]