The length of the perpendicular drawn from the point `(3, -1, 11)` to the line `(x)/(2)=(y-2)/(3)=(z-3)/(4)` is

Options

A. `sqrt(66)`
B. `sqrt(29)`
C. `sqrt(33)`
D. `sqrt(53)`

Answer

To find the length of the perpendicular drawn from the point \( P(3, -1, 11) \) to the line given by the symmetric equations \( \frac{x}{2} = \frac{y-2}{3} = \frac{z-3}{4} \), we will follow these steps:

### Step 1: Parametrize the line
The line can be expressed in parametric form by introducing a parameter \( \lambda \):
- From \( \frac{x}{2} = \lambda \), we have \( x = 2\lambda \).
- From \( \frac{y-2}{3} = \lambda \), we have \( y = 3\lambda + 2 \).
- From \( \frac{z-3}{4} = \lambda \), we have \( z = 4\lambda + 3 \).

Thus, the parametric equations of the line are:
\[ (x, y, z) = (2\lambda, 3\lambda + 2, 4\lambda + 3) \]

### Step 2: Find the direction ratios of the line
The direction ratios of the line can be obtained from the coefficients of \( \lambda \) in the parametric equations:
- Direction ratios: \( (2, 3, 4) \).

### Step 3: Find the vector from point \( P \) to a point on the line
Let \( Q(2\lambda, 3\lambda + 2, 4\lambda + 3) \) be a point on the line. The vector \( \overrightarrow{PQ} \) from point \( P(3, -1, 11) \) to point \( Q \) is given by:
\[ \overrightarrow{PQ} = (2\lambda - 3, (3\lambda + 2) - (-1), (4\lambda + 3) - 11) \]
This simplifies to:
\[ \overrightarrow{PQ} = (2\lambda - 3, 3\lambda + 3, 4\lambda - 8) \]

### Step 4: Set up the perpendicular condition
For \( \overrightarrow{PQ} \) to be perpendicular to the direction ratios of the line \( (2, 3, 4) \), the dot product must be zero:
\[ (2\lambda - 3) \cdot 2 + (3\lambda + 3) \cdot 3 + (4\lambda - 8) \cdot 4 = 0 \]

### Step 5: Expand and simplify the equation
Expanding the dot product:
\[ 4\lambda - 6 + 9\lambda + 9 + 16\lambda - 32 = 0 \]
Combining like terms:
\[ (4\lambda + 9\lambda + 16\lambda) + (-6 + 9 - 32) = 0 \]
This simplifies to:
\[ 29\lambda - 29 = 0 \]
Thus, we find:
\[ \lambda = 1 \]

### Step 6: Find the coordinates of point \( Q \)
Substituting \( \lambda = 1 \) back into the parametric equations:
\[ Q = (2 \cdot 1, 3 \cdot 1 + 2, 4 \cdot 1 + 3) = (2, 5, 7) \]

### Step 7: Calculate the distance \( PQ \)
Now, we can find the distance \( d \) between points \( P(3, -1, 11) \) and \( Q(2, 5, 7) \) using the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]
Substituting the coordinates:
\[ d = \sqrt{(2 - 3)^2 + (5 - (-1))^2 + (7 - 11)^2} \]
Calculating each term:
\[ d = \sqrt{(-1)^2 + (6)^2 + (-4)^2} = \sqrt{1 + 36 + 16} = \sqrt{53} \]

### Final Answer
The length of the perpendicular drawn from the point \( (3, -1, 11) \) to the line is \( \sqrt{53} \).

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