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 barc)|` 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:

\[ |( \hat{d} \cdot \bar{a})(\bar{b} \times \bar{c}) + (\hat{d} \cdot \bar{b})(\bar{c} \times \bar{a}) + (\hat{d} \cdot \bar{c})(\bar{a} \times \bar{b})| \]

Given that \(\hat{d}\) is a unit vector expressed as:

\[ \hat{d} = \lambda (\bar{b} \times \bar{c}) + \mu (\bar{c} \times \bar{a}) + v (\bar{a} \times \bar{b}) \]

### Step 1: Compute \(\hat{d} \cdot \bar{a}\)

Using the properties of the dot product and the cross product, we can compute:

\[ \hat{d} \cdot \bar{a} = \lambda (\bar{b} \times \bar{c}) \cdot \bar{a} + \mu (\bar{c} \times \bar{a}) \cdot \bar{a} + v (\bar{a} \times \bar{b}) \cdot \bar{a} \]

The second and third terms become zero because the dot product of any vector with itself is zero. Thus, we have:

\[ \hat{d} \cdot \bar{a} = \lambda (\bar{b} \times \bar{c}) \cdot \bar{a} \]

### Step 2: Compute \(\hat{d} \cdot \bar{b}\)

Similarly, we compute:

\[ \hat{d} \cdot \bar{b} = \lambda (\bar{b} \times \bar{c}) \cdot \bar{b} + \mu (\bar{c} \times \bar{a}) \cdot \bar{b} + v (\bar{a} \times \bar{b}) \cdot \bar{b} \]

Again, the first and third terms become zero. Thus, we have:

\[ \hat{d} \cdot \bar{b} = \mu (\bar{c} \times \bar{a}) \cdot \bar{b} \]

### Step 3: Compute \(\hat{d} \cdot \bar{c}\)

Now, we compute:

\[ \hat{d} \cdot \bar{c} = \lambda (\bar{b} \times \bar{c}) \cdot \bar{c} + \mu (\bar{c} \times \bar{a}) \cdot \bar{c} + v (\bar{a} \times \bar{b}) \cdot \bar{c} \]

Again, the first and second terms become zero. Thus, we have:

\[ \hat{d} \cdot \bar{c} = v (\bar{a} \times \bar{b}) \cdot \bar{c} \]

### Step 4: Substitute back into the original expression

Now substituting these results back into the original expression:

\[ |( \lambda (\bar{b} \times \bar{c}) \cdot \bar{a})(\bar{b} \times \bar{c}) + (\mu (\bar{c} \times \bar{a}) \cdot \bar{b})(\bar{c} \times \bar{a}) + (v (\bar{a} \times \bar{b}) \cdot \bar{c})(\bar{a} \times \bar{b})| \]

### Step 5: Simplifying the expression

Using the properties of determinants, we can express this as:

\[ = | \lambda \cdot \text{det}(\bar{a}, \bar{b}, \bar{c}) + \mu \cdot \text{det}(\bar{b}, \bar{c}, \bar{a}) + v \cdot \text{det}(\bar{c}, \bar{a}, \bar{b}) | \]

### Step 6: Final result

Since \(\hat{d}\) is a unit vector, the magnitude of this expression simplifies to:

\[ = |\text{det}(\bar{a}, \bar{b}, \bar{c})| \]

Thus, the final answer is:

\[ |\text{det}(\bar{a}, \bar{b}, \bar{c})| \]