What is a vector of unit length orthogonal to both the vectors
`hat(i) + hat(j) + hat(k) and 2 hat(i) + 3 hat(j) - hat(k)`?

Options

A. `(-4hat(i)+3hat(j)-hat(k))/(sqrt(26))`
B. `(-4hat(i)+3hat(j)+hat(k))/(sqrt(26))`
C. `(-3hat(i)+2hat(j)-hat(k))/(sqrt(14))`
D. `(-3hat(i)+2hat(j)+hat(k))/(sqrt(14))`

Explanation

To find a vector of unit length that is orthogonal to both vectors \( \hat{i} + \hat{j} + \hat{k} \) and \( 2\hat{i} + 3\hat{j} - \hat{k} \), we can follow these steps:

### Step 1: Define the Vectors
Let:
- \( \mathbf{A} = \hat{i} + \hat{j} + \hat{k} \)
- \( \mathbf{B} = 2\hat{i} + 3\hat{j} - \hat{k} \)

### Step 2: Compute the Cross Product
To find a vector orthogonal to both \( \mathbf{A} \) and \( \mathbf{B} \), we compute the cross product \( \mathbf{A} \times \mathbf{B} \).

Using the determinant form for the cross product:
\[ \mathbf{A} \times \mathbf{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & 3 & -1 \end{vmatrix} \]

### Step 3: Calculate the Determinant
Expanding the determinant:
\[ \mathbf{A} \times \mathbf{B} = \hat{i} \begin{vmatrix} 1 & 1 \\ 3 & -1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 1 \\ 2 & -1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 1 \\ 2 & 3 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. \( \begin{vmatrix} 1 & 1 \\ 3 & -1 \end{vmatrix} = (1)(-1) - (1)(3) = -1 - 3 = -4 \)
2. \( \begin{vmatrix} 1 & 1 \\ 2 & -1 \end{vmatrix} = (1)(-1) - (1)(2) = -1 - 2 = -3 \)
3. \( \begin{vmatrix} 1 & 1 \\ 2 & 3 \end{vmatrix} = (1)(3) - (1)(2) = 3 - 2 = 1 \)

Putting it all together:
\[ \mathbf{A} \times \mathbf{B} = -4\hat{i} + 3\hat{j} + 1\hat{k} \]

### Step 4: Find the Magnitude of the Cross Product
Now, we calculate the magnitude of the vector \( \mathbf{C} = -4\hat{i} + 3\hat{j} + \hat{k} \):
\[ |\mathbf{C}| = \sqrt{(-4)^2 + (3)^2 + (1)^2} = \sqrt{16 + 9 + 1} = \sqrt{26} \]

### Step 5: Normalize the Vector
To find the unit vector \( \mathbf{U} \) in the direction of \( \mathbf{C} \):
\[ \mathbf{U} = \frac{\mathbf{C}}{|\mathbf{C}|} = \frac{-4\hat{i} + 3\hat{j} + \hat{k}}{\sqrt{26}} \]

### Final Answer
Thus, the vector of unit length orthogonal to both given vectors is:
\[ \mathbf{U} = \frac{-4\hat{i} + 3\hat{j} + \hat{k}}{\sqrt{26}} \]