A unit vector parallel to the intersection of the planes `vecr.(hati-hatj+hatk)=5 andvecr.(2hati+hatj-3hatk)=4` is

Options

A. `(2hati+5hatj-3hatk)/(sqrt38)`
B. `(2hati-5hatj+3hatk)/(sqrt38)`
C. `(-2hati-5hatj-3hatk)/(sqrt38)`
D. `(-2hati+5hatj-3hatk)/(sqrt38)`

Explanation

To find a unit vector parallel to the intersection of the given planes, we will follow these steps:

### Step 1: Identify the normal vectors of the planes
The equations of the planes are given as:
1. \(\vec{r} \cdot (\hat{i} - \hat{j} + \hat{k}) = 5\)
2. \(\vec{r} \cdot (2\hat{i} + \hat{j} - 3\hat{k}) = 4\)

From these equations, we can identify the normal vectors:
- For the first plane, the normal vector \( \vec{n_1} = \hat{i} - \hat{j} + \hat{k} = (1, -1, 1) \)
- For the second plane, the normal vector \( \vec{n_2} = 2\hat{i} + \hat{j} - 3\hat{k} = (2, 1, -3) \)

### Step 2: Compute the cross product of the normal vectors
The direction of the line of intersection of the two planes can be found using the cross product of the normal vectors \( \vec{n_1} \) and \( \vec{n_2} \).

\[ \vec{p} = \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 2 & 1 & -3 \end{vmatrix} \]

Calculating the determinant:

\[ \vec{p} = \hat{i} \left((-1)(-3) - (1)(1)\right) - \hat{j} \left((1)(-3) - (1)(2)\right) + \hat{k} \left((1)(1) - (-1)(2)\right) \]

\[ = \hat{i} (3 - 1) - \hat{j} (-3 - 2) + \hat{k} (1 + 2) \]

\[ = 2\hat{i} + 5\hat{j} + 3\hat{k} \]

### Step 3: Find the modulus of the vector
Now, we need to find the modulus of the vector \( \vec{p} \):

\[ |\vec{p}| = \sqrt{(2)^2 + (5)^2 + (3)^2} = \sqrt{4 + 25 + 9} = \sqrt{38} \]

### Step 4: Find the unit vector
To find the unit vector \( \hat{p} \) in the direction of \( \vec{p} \), we divide \( \vec{p} \) by its modulus:

\[ \hat{p} = \frac{\vec{p}}{|\vec{p}|} = \frac{2\hat{i} + 5\hat{j} + 3\hat{k}}{\sqrt{38}} \]

Thus, the unit vector parallel to the intersection of the planes is:

\[ \hat{p} = \frac{2}{\sqrt{38}} \hat{i} + \frac{5}{\sqrt{38}} \hat{j} + \frac{3}{\sqrt{38}} \hat{k} \]

### Final Answer
The unit vector parallel to the intersection of the planes is:

\[ \hat{p} = \frac{2}{\sqrt{38}} \hat{i} + \frac{5}{\sqrt{38}} \hat{j} + \frac{3}{\sqrt{38}} \hat{k} \]