If the line `(x-4)/(1)=(y-2)/(1)=(z-q)/(p)` lies completely in the plane `2x-4y+z=7`, then the ordered pair (p, q) is

Options

A. (2, 7)
B. (7, 2)
C. (2, 4)
D. (1, 1)

Answer

To solve the problem step by step, we need to analyze the given line and the plane equation.

### Step 1: Identify the line and the plane
The line is given by the equation:
\[ \frac{x-4}{1} = \frac{y-2}{1} = \frac{z-q}{p} \]
This can be interpreted as a parametric line with a point \((4, 2, q)\) and direction ratios \((1, 1, p)\).

The plane is given by the equation:
\[ 2x - 4y + z = 7 \]

### Step 2: Substitute the point into the plane equation
Since the line lies completely in the plane, the point \((4, 2, q)\) must satisfy the plane equation. We substitute \(x = 4\), \(y = 2\), and \(z = q\) into the plane equation:
\[ 2(4) - 4(2) + q = 7 \]
This simplifies to:
\[ 8 - 8 + q = 7 \]
Thus, we find:
\[ q = 7 \]

### Step 3: Find the normal vector of the plane
The normal vector \(\mathbf{n}\) of the plane \(2x - 4y + z = 7\) can be derived from the coefficients of \(x\), \(y\), and \(z\):
\[ \mathbf{n} = (2, -4, 1) \]

### Step 4: Find the direction vector of the line
The direction vector \(\mathbf{b}\) of the line is given by the direction ratios:
\[ \mathbf{b} = (1, 1, p) \]

### Step 5: Use the dot product to find \(p\)
Since the line lies in the plane, the direction vector \(\mathbf{b}\) must be perpendicular to the normal vector \(\mathbf{n}\). Therefore, their dot product must equal zero:
\[ \mathbf{b} \cdot \mathbf{n} = 0 \]
Calculating the dot product:
\[ (1)(2) + (1)(-4) + (p)(1) = 0 \]
This simplifies to:
\[ 2 - 4 + p = 0 \]
Thus:
\[ p - 2 = 0 \implies p = 2 \]

### Step 6: Write the ordered pair \((p, q)\)
Now that we have found \(p\) and \(q\):
\[ (p, q) = (2, 7) \]

### Final Answer
The ordered pair \((p, q)\) is:
\[ \boxed{(2, 7)} \]