What are the direction ratios of the line determined by the planes `x-y+2z=1 and x+y-z=3`?

Options

A. (-1,3,2)
B. (-1,-32)
C. (2,1,3)
D. (2,3,2)

Explanation

To find the direction ratios of the line determined by the intersection of the planes given by the equations \( x - y + 2z = 1 \) and \( x + y - z = 3 \), we can follow these steps:

### Step 1: Write the equations of the planes
The equations of the planes can be rewritten in the standard form:
1. Plane 1: \( x - y + 2z - 1 = 0 \) (let's denote this as \( P_1 \))
2. Plane 2: \( x + y - z - 3 = 0 \) (let's denote this as \( P_2 \))

### Step 2: Identify the normal vectors of the planes
The normal vector of a plane given by the equation \( ax + by + cz = d \) is \( (a, b, c) \).
- For Plane \( P_1 \): The normal vector \( \mathbf{n_1} = (1, -1, 2) \)
- For Plane \( P_2 \): The normal vector \( \mathbf{n_2} = (1, 1, -1) \)

### Step 3: Find the direction ratios of the line of intersection
The direction ratios of the line of intersection of two planes can be found using the cross product of their normal vectors:
\[ \mathbf{d} = \mathbf{n_1} \times \mathbf{n_2} \]

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

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

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

Putting it all together:
\[ \mathbf{d} = \mathbf{i}(-1) - \mathbf{j}(-3) + \mathbf{k}(2) = -\mathbf{i} + 3\mathbf{j} + 2\mathbf{k} \]

Thus, the direction ratios of the line determined by the intersection of the two planes are:
\[ (-1, 3, 2) \]

### Final Answer:
The direction ratios of the line determined by the planes \( x - y + 2z = 1 \) and \( x + y - z = 3 \) are \( (-1, 3, 2) \).

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