The length of the perpendicular (in units) from the point (1, 2, 4) on the straight line `(x-2)/(1)=(y-7)/(2)=(z-3)/(-1)` lies in the interval

Options

A. `(1, (3)/(2))`
B. `(2, 3)`
C. `(0, 2]`
D. `[4, 5)`

Answer

To find the length of the perpendicular from the point \( P(1, 2, 4) \) to the given line represented by the equations \( \frac{x-2}{1} = \frac{y-7}{2} = \frac{z-3}{-1} \), we can follow these steps:

### Step 1: Parametrize the Line
The line can be parametrized using a parameter \( \lambda \). From the given equations, we can express the coordinates of any point \( M \) on the line as:
\[ M(\lambda) = (2 + \lambda, 7 + 2\lambda, 3 - \lambda) \]

### Step 2: Find the Direction Ratios
The direction ratios of the line are given by the coefficients of \( \lambda \) in the parametric equations:
- \( a = 1 \)
- \( b = 2 \)
- \( c = -1 \)

### Step 3: Formulate the Vector PM
The vector \( \overrightarrow{PM} \) from point \( P(1, 2, 4) \) to point \( M(\lambda) \) is given by:
\[ \overrightarrow{PM} = M(\lambda) - P = (2 + \lambda - 1, 7 + 2\lambda - 2, 3 - \lambda - 4) = (\lambda + 1, 2\lambda + 5, -\lambda - 1) \]

### Step 4: Set Up the Perpendicular Condition
For the vector \( \overrightarrow{PM} \) to be perpendicular to the direction ratios of the line, their dot product must be zero:
\[ (\lambda + 1) \cdot 1 + (2\lambda + 5) \cdot 2 + (-\lambda - 1) \cdot (-1) = 0 \]
Expanding this gives:
\[ \lambda + 1 + 4\lambda + 10 + \lambda + 1 = 0 \]
Combining like terms:
\[ 6\lambda + 12 = 0 \]
Thus,
\[ \lambda = -2 \]

### Step 5: Find the Coordinates of Point M
Substituting \( \lambda = -2 \) back into the parametric equations:
\[ M(-2) = (2 - 2, 7 - 4, 3 + 2) = (0, 3, 5) \]

### Step 6: Calculate the Length of PM
Now, we can calculate the length of the perpendicular \( PM \) using the distance formula:
\[ PM = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]
Substituting the coordinates \( P(1, 2, 4) \) and \( M(0, 3, 5) \):
\[ PM = \sqrt{(0 - 1)^2 + (3 - 2)^2 + (5 - 4)^2} \]
Calculating each term:
\[ = \sqrt{(-1)^2 + (1)^2 + (1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \]

### Step 7: Determine the Interval
The numerical value of \( \sqrt{3} \) is approximately \( 1.732 \). We need to find the interval in which this value lies. The intervals provided in the question are:
- \( (0, 2) \)
- \( (1, 3.2) \)
- \( (2, 3) \)

Since \( 1.732 \) is greater than \( 1 \) and less than \( 2 \), the correct interval is \( (1, 2) \).

### Final Answer
The length of the perpendicular from the point \( (1, 2, 4) \) to the line lies in the interval \( (1, 2) \).