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

Options

A. `+- (hati + hatj - 2hatk)/sqrt6`
B. `+- (hati + hatj - hatk)/sqrt3`
C. `+- (hati + hatj + hatk)/sqrt3`
D. `+- hatk`

Explanation

To solve the problem step by step, we will follow the conditions given in the question regarding the vectors \( \vec{a} \), \( \vec{b} \), \( \vec{c} \), and the unit vector \( \hat{d} \).

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

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

### Step 4: Use the scalar triple product condition
The scalar triple product \( [\vec{b}, \vec{c}, \hat{d}] = 0 \):
\[ \begin{vmatrix} \hat{b} & \hat{c} & \hat{d} \\ 0 & 1 & -1 \\ 0 & 0 & 1 \\ \end{vmatrix} = 0 \]
Calculating the determinant:
\[ \begin{vmatrix} 0 & 1 & -1 \\ 0 & 0 & 1 \\ x & y & z \\ \end{vmatrix} = 0 \]
Expanding along the first row:
\[ 0 \cdot \begin{vmatrix} 0 & 1 \\ y & z \end{vmatrix} - 1 \cdot \begin{vmatrix} 0 & 1 \\ x & z \end{vmatrix} + (-1) \cdot \begin{vmatrix} 0 & 1 \\ x & y \end{vmatrix} = 0 \]
This simplifies to:
\[ -x + y = 0 \quad \Rightarrow \quad x + y + z = 0 \quad \text{(Equation 3)} \]

### Step 5: Substitute Equation 2 into Equation 3
Substituting \( y = x \) into Equation 3:
\[ x + x + z = 0 \quad \Rightarrow \quad 2x + z = 0 \quad \Rightarrow \quad z = -2x \quad \text{(Equation 4)} \]

### Step 6: Substitute Equations 2 and 4 into Equation 1
Substituting \( y = x \) and \( z = -2x \) into Equation 1:
\[ x^2 + x^2 + (-2x)^2 = 1 \]
This simplifies to:
\[ x^2 + x^2 + 4x^2 = 1 \quad \Rightarrow \quad 6x^2 = 1 \quad \Rightarrow \quad x^2 = \frac{1}{6} \]
Thus:
\[ x = \pm \frac{1}{\sqrt{6}}, \quad y = \pm \frac{1}{\sqrt{6}}, \quad z = -2x = \mp \frac{2}{\sqrt{6}} \]

### Step 7: Write the final expression for \( \hat{d} \)
Thus, the unit vector \( \hat{d} \) is:
\[ \hat{d} = \pm \frac{1}{\sqrt{6}} \hat{i} + \pm \frac{1}{\sqrt{6}} \hat{j} - \frac{2}{\sqrt{6}} \hat{k} \]
This can be expressed as:
\[ \hat{d} = \frac{1}{\sqrt{6}} \hat{i} + \frac{1}{\sqrt{6}} \hat{j} - \frac{2}{\sqrt{6}} \hat{k} \quad \text{or} \quad \hat{d} = -\frac{1}{\sqrt{6}} \hat{i} - \frac{1}{\sqrt{6}} \hat{j} + \frac{2}{\sqrt{6}} \hat{k} \]

### Final Answer
The unit vector \( \hat{d} \) can be expressed as:
\[ \hat{d} = \frac{1}{\sqrt{6}} \hat{i} + \frac{1}{\sqrt{6}} \hat{j} - \frac{2}{\sqrt{6}} \hat{k} \quad \text{or} \quad \hat{d} = -\frac{1}{\sqrt{6}} \hat{i} - \frac{1}{\sqrt{6}} \hat{j} + \frac{2}{\sqrt{6}} \hat{k} \]