If `veca =hati + hatj, vecb = hati - hatj + hatk` and `vecc` is a unit vector `bot` to the vector `veca` and coplanar with `veca` and `vecb`, then a unit vector `vecd` is perpendicular to both `veca` and `vecc` is:

Options

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

Answer

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

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

### Step 2: Assume the form of vector \( \vec{c} \)
Let:
\[ \vec{c} = x \hat{i} + y \hat{j} + z \hat{k} \]
Since \( \vec{c} \) is perpendicular to \( \vec{a} \), we have:
\[ \vec{c} \cdot \vec{a} = 0 \]
This gives us:
\[ x + y = 0 \quad \text{(1)} \]
From equation (1), we can express \( y \) in terms of \( x \):
\[ y = -x \]

### Step 3: Check coplanarity with \( \vec{a} \) and \( \vec{b} \)
The condition for coplanarity is that the scalar triple product \( \vec{c}, \vec{a}, \vec{b} \) must be zero:
\[ \vec{c} \cdot (\vec{a} \times \vec{b}) = 0 \]

### Step 4: Calculate \( \vec{a} \times \vec{b} \)
Using the determinant method:
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 0 \\ 1 & -1 & 1 \end{vmatrix} \]
Calculating the determinant:
\[ = \hat{i} \begin{vmatrix} 1 & 0 \\ -1 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 0 \\ 1 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix} \]
\[ = \hat{i} (1 - 0) - \hat{j} (1 - 0) + \hat{k} (-1 - 1) \]
\[ = \hat{i} - \hat{j} - 2\hat{k} \]
Thus:
\[ \vec{a} \times \vec{b} = \hat{i} - \hat{j} - 2\hat{k} \]

### Step 5: Set up the equation for \( \vec{c} \)
Now we have:
\[ \vec{c} \cdot (\hat{i} - \hat{j} - 2\hat{k}) = 0 \]
Substituting \( \vec{c} = x \hat{i} - x \hat{j} + z \hat{k} \):
\[ x - (-x) - 2z = 0 \]
This simplifies to:
\[ 2x - 2z = 0 \quad \text{(2)} \]
From equation (2):
\[ x = z \]

### Step 6: Express \( \vec{c} \)
Now substituting \( z = x \) into \( \vec{c} \):
\[ \vec{c} = x \hat{i} - x \hat{j} + x \hat{k} = x (\hat{i} - \hat{j} + \hat{k}) \]

### Step 7: Normalize \( \vec{c} \)
To make \( \vec{c} \) a unit vector:
\[ \|\vec{c}\| = 1 \Rightarrow x \sqrt{1^2 + (-1)^2 + 1^2} = 1 \]
\[ x \sqrt{3} = 1 \Rightarrow x = \frac{1}{\sqrt{3}} \]
Thus:
\[ \vec{c} = \frac{1}{\sqrt{3}} (\hat{i} - \hat{j} + \hat{k}) \]

### Step 8: Find \( \vec{d} \)
Now, \( \vec{d} \) is perpendicular to both \( \vec{a} \) and \( \vec{c} \), so:
\[ \vec{d} \parallel \vec{a} \times \vec{c} \]
Calculating \( \vec{a} \times \vec{c} \):
\[ \vec{a} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 0 \\ \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \end{vmatrix} \]
Calculating the determinant:
\[ = \hat{i} \begin{vmatrix} 1 & 0 \\ -\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 0 \\ \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 1 \\ \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{3}} \end{vmatrix} \]
Calculating each determinant:
\[ = \hat{i} \left( \frac{1}{\sqrt{3}} \right) - \hat{j} \left( \frac{1}{\sqrt{3}} \right) + \hat{k} \left( -\frac{1}{\sqrt{3}} - \frac{1}{\sqrt{3}} \right) \]
\[ = \frac{1}{\sqrt{3}} \hat{i} - \frac{1}{\sqrt{3}} \hat{j} - \frac{2}{\sqrt{3}} \hat{k} \]

### Step 9: Normalize \( \vec{d} \)
Let \( \vec{d} = \lambda \left( \frac{1}{\sqrt{3}} \hat{i} - \frac{1}{\sqrt{3}} \hat{j} - \frac{2}{\sqrt{3}} \hat{k} \right) \) and set its magnitude to 1:
\[ \|\vec{d}\| = \lambda \sqrt{\left( \frac{1}{\sqrt{3}} \right)^2 + \left( -\frac{1}{\sqrt{3}} \right)^2 + \left( -\frac{2}{\sqrt{3}} \right)^2} = 1 \]
\[ \lambda \sqrt{\frac{1}{3} + \frac{1}{3} + \frac{4}{3}} = 1 \]
\[ \lambda \sqrt{2} = 1 \Rightarrow \lambda = \frac{1}{\sqrt{2}} \]
Thus:
\[ \vec{d} = \frac{1}{\sqrt{2}} \left( \frac{1}{\sqrt{3}} \hat{i} - \frac{1}{\sqrt{3}} \hat{j} - \frac{2}{\sqrt{3}} \hat{k} \right) \]
This simplifies to:
\[ \vec{d} = \frac{1}{\sqrt{6}} \hat{i} - \frac{1}{\sqrt{6}} \hat{j} - \frac{2}{\sqrt{6}} \hat{k} \]

### Final Answer
The unit vector \( \vec{d} \) that is perpendicular to both \( \vec{a} \) and \( \vec{c} \) is:
\[ \vec{d} = \frac{1}{\sqrt{6}} \hat{i} - \frac{1}{\sqrt{6}} \hat{j} - \frac{2}{\sqrt{6}} \hat{k} \]