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 find the magnitude of the vector \(\vec{c}\) that is perpendicular to both \(\vec{a}\) and \(\vec{b}\), we will follow these steps:

### Step 1: Define the vectors
Given:
\[ \vec{a} = \hat{i} + \hat{j} - \hat{k} \]
\[ \vec{b} = 2\hat{i} - \hat{j} + \hat{k} \]

### Step 2: Calculate the cross product \(\vec{c} = \vec{a} \times \vec{b}\)
To find a vector \(\vec{c}\) that is perpendicular to both \(\vec{a}\) and \(\vec{b}\), we compute the cross product \(\vec{a} \times \vec{b}\).

The cross product is given by the determinant:
\[ \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -1 \\ 2 & -1 & 1 \end{vmatrix} \]

### Step 3: Calculate the determinant
Expanding the determinant:
\[ \vec{c} = \hat{i} \begin{vmatrix} 1 & -1 \\ -1 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & -1 \\ 2 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 1 \\ 2 & -1 \end{vmatrix} \]

Calculating the 2x2 determinants:
1. \(\begin{vmatrix} 1 & -1 \\ -1 & 1 \end{vmatrix} = 1 \cdot 1 - (-1) \cdot (-1) = 1 - 1 = 0\)
2. \(\begin{vmatrix} 1 & -1 \\ 2 & 1 \end{vmatrix} = 1 \cdot 1 - (-1) \cdot 2 = 1 + 2 = 3\)
3. \(\begin{vmatrix} 1 & 1 \\ 2 & -1 \end{vmatrix} = 1 \cdot (-1) - 1 \cdot 2 = -1 - 2 = -3\)

Putting it all together:
\[ \vec{c} = \hat{i}(0) - \hat{j}(3) + \hat{k}(-3) = -3\hat{j} - 3\hat{k} \]
Thus,
\[ \vec{c} = 0\hat{i} - 3\hat{j} - 3\hat{k} \]

### Step 4: Find the magnitude of \(\vec{c}\)
The magnitude of \(\vec{c}\) is given by:
\[ |\vec{c}| = \sqrt{(0)^2 + (-3)^2 + (-3)^2} = \sqrt{0 + 9 + 9} = \sqrt{18} = 3\sqrt{2} \]

### Final Answer
\[ |\vec{c}| = 3\sqrt{2} \]