Let `veca=hati+hatj-hatk` and `vecb=2hati-hatj+hatk`. If `vecc` is a non - zero vector perpendicular to both `veca and vecb`, such that its component along x, y, z axes are rational numbers, then `|vecc|` is

Options

A. an integer
B. an irrational number
C. a rational number
D. None of these

Answer

To solve the problem step by step, we need to find the magnitude of the vector \(\vec{c}\) which is perpendicular to both \(\vec{a}\) and \(\vec{b}\) and has rational components.

### Step 1: Define the vectors
We have:
\[ \vec{a} = \hat{i} + \hat{j} - \hat{k} \]
\[ \vec{b} = 2\hat{i} - \hat{j} + \hat{k} \]
Let \(\vec{c} = x\hat{i} + y\hat{j} + z\hat{k}\), where \(x\), \(y\), and \(z\) are rational numbers.

### Step 2: Set up the dot product equations
Since \(\vec{c}\) is perpendicular to \(\vec{a}\), we have:
\[ \vec{a} \cdot \vec{c} = 0 \]
This gives us:
\[ (x + y - z) = 0 \quad \text{(Equation 1)} \]

Similarly, since \(\vec{c}\) is also perpendicular to \(\vec{b}\), we have:
\[ \vec{b} \cdot \vec{c} = 0 \]
This gives us:
\[ (2x - y + z) = 0 \quad \text{(Equation 2)} \]

### Step 3: Solve the equations
From Equation 1:
\[ z = x + y \]

Substituting \(z\) into Equation 2:
\[ 2x - y + (x + y) = 0 \]
This simplifies to:
\[ 3x = 0 \]
Thus, we find:
\[ x = 0 \]

### Step 4: Substitute back to find \(y\) and \(z\)
Using \(x = 0\) in Equation 1:
\[ 0 + y - z = 0 \implies z = y \]

### Step 5: Express \(\vec{c}\)
Now, substituting \(x = 0\) and \(z = y\):
\[ \vec{c} = 0\hat{i} + y\hat{j} + y\hat{k} = y(\hat{j} + \hat{k}) \]

### Step 6: Find the magnitude of \(\vec{c}\)
The magnitude of \(\vec{c}\) is given by:
\[ |\vec{c}| = \sqrt{(0)^2 + (y)^2 + (y)^2} = \sqrt{2y^2} = |y|\sqrt{2} \]

### Step 7: Determine the nature of \(|\vec{c}|\)
Since \(y\) is a rational number, \(|y|\sqrt{2}\) is irrational (as the product of a rational number and an irrational number is irrational).

### Conclusion
Thus, the magnitude \(|\vec{c}|\) is irrational.

### Final Answer
\(|\vec{c}| = |y|\sqrt{2}\) (which is irrational).