Let `veca=hati-hatj,vecb=hatj-hatk, vecc=hatk-hati`. If `hatd` is a unit vector such that `veca.hatd=0=[(vecb ,vecc, hatd)]`, then `hatd` equals

Options

A. `+-(hati+hatj-2hatk)/(sqrt(6))`
B. `+-(hati+hatj-hatk)/(sqrt(3))`
C. `+-(hati+hatj+hatk)/(sqrt(3))`
D. `+-hatk`

Explanation

To find the unit vector \(\hat{d}\) such that \(\vec{a} \cdot \hat{d} = 0\) and \([\vec{b}, \vec{c}, \hat{d}] = 0\), we will follow these steps:

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

### Step 2: Express \(\hat{d}\) as a general unit vector
Let:
\[ \hat{d} = x \hat{i} + y \hat{j} + z \hat{k} \]
Since \(\hat{d}\) is a unit vector, we have:
\[ x^2 + y^2 + z^2 = 1 \quad \text{(Equation 1)} \]

### Step 3: Use the dot product condition
The condition \(\vec{a} \cdot \hat{d} = 0\) gives:
\[ (\hat{i} - \hat{j}) \cdot (x \hat{i} + y \hat{j} + z \hat{k}) = 0 \]
Calculating the dot product:
\[ x - y = 0 \implies x = y \quad \text{(Equation 2)} \]

### Step 4: Use the scalar triple product condition
The condition \([\vec{b}, \vec{c}, \hat{d}] = 0\) implies that the vectors \(\vec{b}\), \(\vec{c}\), and \(\hat{d}\) are coplanar. This can be represented using the determinant:
\[ \begin{vmatrix} 0 & 1 & -1 \\ -1 & 0 & 1 \\ x & y & z \end{vmatrix} = 0 \]
Calculating the determinant:
\[ 0 \cdot (0 \cdot z - 1 \cdot y) - 1 \cdot (-1 \cdot z - 1 \cdot x) + (-1) \cdot (0 \cdot y - (-1) \cdot x) = 0 \]
This simplifies to:
\[ -z - x + x = 0 \implies -z = 0 \implies z = 0 \quad \text{(Equation 3)} \]

### Step 5: Substitute \(z\) into Equation 1
Substituting \(z = 0\) into Equation 1:
\[ x^2 + y^2 + 0^2 = 1 \implies x^2 + y^2 = 1 \]
Using Equation 2, where \(x = y\):
\[ x^2 + x^2 = 1 \implies 2x^2 = 1 \implies x^2 = \frac{1}{2} \implies x = \pm \frac{1}{\sqrt{2}} \]
Thus, \(y = \pm \frac{1}{\sqrt{2}}\).

### Step 6: Determine \(\hat{d}\)
Since \(z = 0\), we have:
\[ \hat{d} = x \hat{i} + y \hat{j} + 0 \hat{k} = \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j} \quad \text{or} \quad \hat{d} = -\frac{1}{\sqrt{2}} \hat{i} - \frac{1}{\sqrt{2}} \hat{j} \]
Thus, the possible unit vectors are:
\[ \hat{d} = \frac{1}{\sqrt{2}} (\hat{i} + \hat{j}) \quad \text{or} \quad \hat{d} = -\frac{1}{\sqrt{2}} (\hat{i} + \hat{j}) \]

### Final Answer
The unit vector \(\hat{d}\) can be expressed as:
\[ \hat{d} = \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j} \]
or
\[ \hat{d} = -\frac{1}{\sqrt{2}} \hat{i} - \frac{1}{\sqrt{2}} \hat{j} \]