A vector parallel to the line of intersection of the planes `overset(to)( r) (3 hat(i) - hat(j) + hat(k) )=5 and overset(to) (r ) (hat(i) +4 hat(j) - 2 hat(k) )=3` is

Options

A. `2hat(i) + 7 hat(j) -13 hat(k)`
B. `2hat(i) - 7hat(i) + 13 hat(k)`
C. `-2hat(i) + 7 hat(j) + 13 hat(k)`
D. `2hat(i) + 7 hat(j) + 13 hat(k)`

Answer

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

### Step 1: Write the equations of the planes in vector form
The equations of the planes are given as:
1. \( 3\hat{i} - \hat{j} + \hat{k} = 5 \)
2. \( \hat{i} + 4\hat{j} - 2\hat{k} = 3 \)

We can express these equations in the standard form:
- Plane 1: \( 3x - y + z = 5 \)
- Plane 2: \( x + 4y - 2z = 3 \)

### Step 2: Identify the normal vectors of the planes
The normal vector of a plane \( Ax + By + Cz = D \) is given by \( \vec{n} = A\hat{i} + B\hat{j} + C\hat{k} \).

For Plane 1, the normal vector \( \vec{n_1} \) is:
\[ \vec{n_1} = 3\hat{i} - \hat{j} + \hat{k} \]

For Plane 2, the normal vector \( \vec{n_2} \) is:
\[ \vec{n_2} = \hat{i} + 4\hat{j} - 2\hat{k} \]

### Step 3: Find the direction vector of the line of intersection
The direction vector \( \vec{d} \) of the line of intersection of the two planes can be found by taking the cross product of the normal vectors \( \vec{n_1} \) and \( \vec{n_2} \).

\[ \vec{d} = \vec{n_1} \times \vec{n_2} \]

Calculating the cross product:
\[ \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 1 \\ 1 & 4 & -2 \end{vmatrix} \]

### Step 4: Calculate the determinant
Using the determinant formula:
\[ \vec{d} = \hat{i} \begin{vmatrix} -1 & 1 \\ 4 & -2 \end{vmatrix} - \hat{j} \begin{vmatrix} 3 & 1 \\ 1 & -2 \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & -1 \\ 1 & 4 \end{vmatrix} \]

Calculating each of these determinants:
1. For \( \hat{i} \):
\[ \begin{vmatrix} -1 & 1 \\ 4 & -2 \end{vmatrix} = (-1)(-2) - (1)(4) = 2 - 4 = -2 \]

2. For \( \hat{j} \):
\[ \begin{vmatrix} 3 & 1 \\ 1 & -2 \end{vmatrix} = (3)(-2) - (1)(1) = -6 - 1 = -7 \]

3. For \( \hat{k} \):
\[ \begin{vmatrix} 3 & -1 \\ 1 & 4 \end{vmatrix} = (3)(4) - (-1)(1) = 12 + 1 = 13 \]

### Step 5: Combine the results
Putting it all together:
\[ \vec{d} = -2\hat{i} + 7\hat{j} + 13\hat{k} \]

### Final Answer
A vector parallel to the line of intersection of the planes is:
\[ \vec{d} = -2\hat{i} + 7\hat{j} + 13\hat{k} \]