Let `veca=hati+hatj-hatk, vecb=hati-hatj+hatk` and `vecc` be a unit vector perpendicular to `veca` and coplanar with `veca` and `vecb`, then it is given by

Options

A. `1/(sqrt(6))(2hati-hatj+hatjk)`
B. `1/(sqrt(2))(hatj+hatk)`
C. `1/(sqrt(6))(hati-2hatj+hatk)`
D. `1/2(hatj-hatk)`

Explanation

To solve the problem, we need to find a unit vector \( \vec{c} \) that is perpendicular to \( \vec{a} \) and coplanar with \( \vec{a} \) and \( \vec{b} \).

Given:
\[ \vec{a} = \hat{i} + \hat{j} - \hat{k} \]
\[ \vec{b} = \hat{i} - \hat{j} + \hat{k} \]
Let \( \vec{c} = x \hat{i} + y \hat{j} + z \hat{k} \).

### Step 1: Condition for Coplanarity
For the vectors \( \vec{a}, \vec{b}, \vec{c} \) to be coplanar, the scalar triple product must be zero:
\[ \vec{a} \cdot (\vec{b} \times \vec{c}) = 0 \]
This can be expressed as the determinant of the matrix formed by the components of the vectors:
\[ \begin{vmatrix} 1 & 1 & -1 \\ 1 & -1 & 1 \\ x & y & z \end{vmatrix} = 0 \]

### Step 2: Calculate the Determinant
Calculating the determinant:
\[ = 1 \begin{vmatrix} -1 & 1 \\ y & z \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ x & z \end{vmatrix} - 1 \begin{vmatrix} 1 & -1 \\ x & y \end{vmatrix} \]
Calculating each of these 2x2 determinants:
\[ = 1((-1)z - 1y) - 1(1z - 1x) - 1(1y + 1x) \]
\[ = -z - y - z + x - y - x \]
\[ = -2z - 2y = 0 \]
Thus, we have:
\[ z + y = 0 \quad \text{(Equation 1)} \]

### Step 3: Condition for Perpendicularity
Since \( \vec{c} \) is perpendicular to \( \vec{a} \):
\[ \vec{a} \cdot \vec{c} = 0 \]
This gives:
\[ 1 \cdot x + 1 \cdot y - 1 \cdot z = 0 \]
Substituting \( z = -y \) from Equation 1:
\[ x + y + y = 0 \]
\[ x + 2y = 0 \quad \text{(Equation 2)} \]

### Step 4: Unit Vector Condition
Since \( \vec{c} \) is a unit vector:
\[ \sqrt{x^2 + y^2 + z^2} = 1 \]
Substituting \( z = -y \) and \( x = -2y \):
\[ \sqrt{(-2y)^2 + y^2 + (-y)^2} = 1 \]
\[ \sqrt{4y^2 + y^2 + y^2} = 1 \]
\[ \sqrt{6y^2} = 1 \]
\[ \sqrt{6} |y| = 1 \implies |y| = \frac{1}{\sqrt{6}} \]
Thus, \( y = \frac{1}{\sqrt{6}} \) or \( y = -\frac{1}{\sqrt{6}} \).

### Step 5: Finding \( x \) and \( z \)
Using \( y = \frac{1}{\sqrt{6}} \):
\[ x = -2y = -2 \cdot \frac{1}{\sqrt{6}} = -\frac{2}{\sqrt{6}} \]
\[ z = -y = -\frac{1}{\sqrt{6}} \]
Thus, we have:
\[ \vec{c} = -\frac{2}{\sqrt{6}} \hat{i} + \frac{1}{\sqrt{6}} \hat{j} - \frac{1}{\sqrt{6}} \hat{k} \]

### Final Result
Factoring out \( \frac{1}{\sqrt{6}} \):
\[ \vec{c} = \frac{1}{\sqrt{6}} \left(-2 \hat{i} + \hat{j} - \hat{k}\right) \]