If `vecb and vecc` are any two mutually perpendicular unit vectors and `veca` is any vector, then `(veca.vecb)vecb+(veca.vecc)vecc+(veca.(vecbxxvecc))/(|vecbxxvecc|^2)(vecbxxvecc)=` (A) 0 (B) `veca (C) `veca /2` (D) `2veca`

Explanation

To solve the given problem, we need to evaluate the expression:

\[ (\vec{a} \cdot \vec{b}) \vec{b} + (\vec{a} \cdot \vec{c}) \vec{c} + \frac{\vec{a} \cdot (\vec{b} \times \vec{c})}{|\vec{b} \times \vec{c}|^2} (\vec{b} \times \vec{c}) \]

where \(\vec{b}\) and \(\vec{c}\) are mutually perpendicular unit vectors, and \(\vec{a}\) is any vector.

### Step 1: Define the vectors
Assume:
- \(\vec{b} = \hat{j}\) (unit vector in the y-direction)
- \(\vec{c} = \hat{k}\) (unit vector in the z-direction)

### Step 2: Calculate the cross product \(\vec{b} \times \vec{c}\)
Using the right-hand rule for cross products:
\[ \vec{b} \times \vec{c} = \hat{j} \times \hat{k} = \hat{i} \]
Thus, \(\vec{b} \times \vec{c} = \hat{i}\).

### Step 3: Calculate the magnitude of \(\vec{b} \times \vec{c}\)
\[ |\vec{b} \times \vec{c}| = |\hat{i}| = 1 \]
Therefore, \(|\vec{b} \times \vec{c}|^2 = 1^2 = 1\).

### Step 4: Express vector \(\vec{a}\)
Let \(\vec{a} = x \hat{i} + y \hat{j} + z \hat{k}\).

### Step 5: Calculate \(\vec{a} \cdot \vec{b}\)
\[ \vec{a} \cdot \vec{b} = (x \hat{i} + y \hat{j} + z \hat{k}) \cdot \hat{j} = y \]

### Step 6: Calculate \(\vec{a} \cdot \vec{c}\)
\[ \vec{a} \cdot \vec{c} = (x \hat{i} + y \hat{j} + z \hat{k}) \cdot \hat{k} = z \]

### Step 7: Calculate \(\vec{a} \cdot (\vec{b} \times \vec{c})\)
\[ \vec{a} \cdot (\vec{b} \times \vec{c}) = (x \hat{i} + y \hat{j} + z \hat{k}) \cdot \hat{i} = x \]

### Step 8: Substitute values into the expression
Now substituting the values we calculated:
\[ (\vec{a} \cdot \vec{b}) \vec{b} = y \hat{j} \]
\[ (\vec{a} \cdot \vec{c}) \vec{c} = z \hat{k} \]
\[ \frac{\vec{a} \cdot (\vec{b} \times \vec{c})}{|\vec{b} \times \vec{c}|^2} (\vec{b} \times \vec{c}) = \frac{x}{1} \hat{i} = x \hat{i} \]

### Step 9: Combine all parts
Now we combine all parts:
\[ y \hat{j} + z \hat{k} + x \hat{i} = x \hat{i} + y \hat{j} + z \hat{k} = \vec{a} \]

### Conclusion
Thus, the expression simplifies to \(\vec{a}\).

The correct answer is (B) \(\vec{a}\).

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