A vector of magnitude 5 and perpendicular to `hat(i) - 2 hat(j) + hat(k)` and `2 hat(i) + hat(j) - 3 hat(k)` is

Options

A. `(5 sqrt(3) ( hat(i) + hat(j) + hat(k)))/( 3)`
B. `(5 sqrt(3) ( hat(i) - hat(j) + hat(k)))/( 3)`
C. `(5 sqrt(3) ( hat(i) - hat(j) - hat(k)))/( 3)`
D. `(5 sqrt(3) ( hat(i) + hat(j) - hat(k)))/( sqrt(3))`

Answer

To solve the problem of finding a vector of magnitude 5 that is perpendicular to the vectors \( \hat{i} - 2\hat{j} + \hat{k} \) and \( 2\hat{i} + \hat{j} - 3\hat{k} \), we can follow these steps:

### Step 1: Define the vectors
Let:
\[ \mathbf{a} = \hat{i} - 2\hat{j} + \hat{k} \]
\[ \mathbf{b} = 2\hat{i} + \hat{j} - 3\hat{k} \]

### Step 2: Calculate the cross product
To find a vector that is perpendicular to both \( \mathbf{a} \) and \( \mathbf{b} \), we compute the cross product \( \mathbf{c} = \mathbf{a} \times \mathbf{b} \).

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

### Step 3: Compute the determinant
Calculating the determinant:
\[ \mathbf{c} = \hat{i} \begin{vmatrix} -2 & 1 \\ 1 & -3 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 1 \\ 2 & -3 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & -2 \\ 2 & 1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. \( \begin{vmatrix} -2 & 1 \\ 1 & -3 \end{vmatrix} = (-2)(-3) - (1)(1) = 6 - 1 = 5 \)
2. \( \begin{vmatrix} 1 & 1 \\ 2 & -3 \end{vmatrix} = (1)(-3) - (1)(2) = -3 - 2 = -5 \)
3. \( \begin{vmatrix} 1 & -2 \\ 2 & 1 \end{vmatrix} = (1)(1) - (-2)(2) = 1 + 4 = 5 \)

Putting it all together:
\[ \mathbf{c} = 5\hat{i} + 5\hat{j} + 5\hat{k} \]

### Step 4: Find the magnitude of \( \mathbf{c} \)
The magnitude of \( \mathbf{c} \) is:
\[ |\mathbf{c}| = \sqrt{5^2 + 5^2 + 5^2} = \sqrt{25 + 25 + 25} = \sqrt{75} = 5\sqrt{3} \]

### Step 5: Find the unit vector in the direction of \( \mathbf{c} \)
The unit vector \( \mathbf{u} \) in the direction of \( \mathbf{c} \) is:
\[ \mathbf{u} = \frac{\mathbf{c}}{|\mathbf{c}|} = \frac{5\hat{i} + 5\hat{j} + 5\hat{k}}{5\sqrt{3}} = \frac{1}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k}) \]

### Step 6: Scale the unit vector to the desired magnitude
To find a vector of magnitude 5 in the direction of \( \mathbf{u} \):
\[ \mathbf{d} = 5 \cdot \mathbf{u} = 5 \cdot \frac{1}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k}) = \frac{5}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k}) \]

### Final Result
Thus, the vector of magnitude 5 that is perpendicular to both given vectors is:
\[ \mathbf{d} = \frac{5}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k}) \]