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:
\[ \vec{a} = \hat{i} + \hat{j} - \hat{k} \]
\[ \vec{b} = \hat{i} - \hat{j} + \hat{k} \]

### Step 1: Define the vector \( \vec{c} \)
Let \( \vec{c} = x \hat{i} + y \hat{j} + z \hat{k} \).

### Step 2: Use the condition of perpendicularity
Since \( \vec{c} \) is perpendicular to \( \vec{a} \), we have:
\[ \vec{c} \cdot \vec{a} = 0 \]
Calculating the dot product:
\[ (x \hat{i} + y \hat{j} + z \hat{k}) \cdot (\hat{i} + \hat{j} - \hat{k}) = x + y - z = 0 \]
This gives us our first equation:
\[ x + y - z = 0 \quad \text{(1)} \]

### Step 3: Use the condition of coplanarity
Since \( \vec{a} \), \( \vec{b} \), and \( \vec{c} \) are coplanar, the scalar triple product must be zero:
\[ \vec{a} \cdot (\vec{b} \times \vec{c}) = 0 \]
Using the determinant method, we can express this as:
\[ \begin{vmatrix} 1 & 1 & -1 \\ 1 & -1 & 1 \\ x & y & z \end{vmatrix} = 0 \]
Calculating the determinant:
\[ 1 \cdot \begin{vmatrix} -1 & 1 \\ y & z \end{vmatrix} - 1 \cdot \begin{vmatrix} 1 & 1 \\ x & z \end{vmatrix} - 1 \cdot \begin{vmatrix} 1 & -1 \\ x & y \end{vmatrix} = 0 \]
Expanding this:
\[ (-1)(z) - (1)(z - x) - (1)(y + x) = 0 \]
This simplifies to:
\[ -z - z + x - y - x = 0 \]
Thus, we have:
\[ -2z + y = 0 \quad \text{(2)} \]

### Step 4: Solve the equations
From equation (2), we can express \( y \) in terms of \( z \):
\[ y = 2z \]
Substituting \( y = 2z \) into equation (1):
\[ x + 2z - z = 0 \]
This simplifies to:
\[ x + z = 0 \implies x = -z \]

### Step 5: Express \( \vec{c} \) in terms of \( z \)
Now we can express \( \vec{c} \):
\[ \vec{c} = -z \hat{i} + 2z \hat{j} + z \hat{k} = z(-\hat{i} + 2\hat{j} + \hat{k}) \]

### Step 6: Normalize \( \vec{c} \) to make it a unit vector
To make \( \vec{c} \) a unit vector, we need to find the magnitude:
\[ \|\vec{c}\| = z \sqrt{(-1)^2 + 2^2 + 1^2} = z \sqrt{1 + 4 + 1} = z \sqrt{6} \]
Setting this equal to 1 for a unit vector:
\[ z \sqrt{6} = 1 \implies z = \frac{1}{\sqrt{6}} \]

### Step 7: Substitute \( z \) back to find \( \vec{c} \)
Substituting \( z \) back into \( \vec{c} \):
\[ \vec{c} = \frac{1}{\sqrt{6}} \left(-\hat{i} + 2\hat{j} + \hat{k}\right) \]

Thus, the required unit vector \( \vec{c} \) is:
\[ \vec{c} = \frac{1}{\sqrt{6}} (-\hat{i} + 2\hat{j} + \hat{k}) \]