What are the direction rations of the line of intersection of the planes `x=3z+4 and y=2z-3`?

Options

A. `ltlt1,2,3gtgt`
B. `ltlt2,1,3gtgt`
C. `ltlt3,2,1gtgt`
D. `ltlt1,3,2gtgt`

Answer

To find the direction ratios of the line of intersection of the planes given by the equations \( x = 3z + 4 \) and \( y = 2z - 3 \), we can follow these steps:

### Step 1: Write the equations of the planes in standard form
The equations of the planes can be rewritten as:
1. For the first plane:
\[ x - 3z - 4 = 0 \quad \text{(Plane 1)} \]
2. For the second plane:
\[ y - 2z + 3 = 0 \quad \text{(Plane 2)} \]

### Step 2: Identify the normal vectors of the planes
From the standard form of the equations, we can identify the normal vectors:
- The normal vector of Plane 1 is \( \mathbf{n_1} = (1, 0, -3) \).
- The normal vector of Plane 2 is \( \mathbf{n_2} = (0, 1, -2) \).

### Step 3: Set up the condition for perpendicularity
The direction ratios of the line of intersection of the planes will be perpendicular to the normal vectors of both planes. Therefore, if we denote the direction ratios of the line as \( (A, B, C) \), we can use the dot product to set up the following equations:
1. For Plane 1:
\[ A \cdot 1 + B \cdot 0 + C \cdot (-3) = 0 \implies A - 3C = 0 \quad \text{(Equation 1)} \]
2. For Plane 2:
\[ A \cdot 0 + B \cdot 1 + C \cdot (-2) = 0 \implies B - 2C = 0 \quad \text{(Equation 2)} \]

### Step 4: Solve the equations
From Equation 1:
\[ A = 3C \]
From Equation 2:
\[ B = 2C \]

### Step 5: Express direction ratios in terms of a parameter
Let \( C = k \) (where \( k \) is a parameter), then:
- \( A = 3k \)
- \( B = 2k \)
- \( C = k \)

Thus, the direction ratios can be expressed as:
\[ (A, B, C) = (3k, 2k, k) = k(3, 2, 1) \]

### Step 6: Choose a convenient value for \( k \)
For simplicity, we can choose \( k = 1 \):
\[ (A, B, C) = (3, 2, 1) \]

### Final Answer
The direction ratios of the line of intersection of the planes are \( (3, 2, 1) \).

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