To find the length of the perpendicular from the point \( P(2, -3, 1) \) to the line given by the symmetric equations \( \frac{x+1}{2} = \frac{y-3}{3} = \frac{z+2}{-1} \), we can follow these steps:
### Step 1: Parameterize the Line
The symmetric equations can be expressed in terms of a parameter \( \lambda \):
\[
\frac{x+1}{2} = \frac{y-3}{3} = \frac{z+2}{-1} = \lambda
\]
From this, we can derive the parametric equations for the line:
\[
x = 2\lambda - 1, \quad y = 3\lambda + 3, \quad z = -\lambda - 2
\]
### Step 2: Find Direction Ratios of the Line
The direction ratios of the line can be extracted from the coefficients of \( \lambda \):
\[
\text{Direction ratios} = (2, 3, -1)
\]
### Step 3: Find the Point on the Line Closest to Point P
Let the point on the line corresponding to the parameter \( \lambda \) be \( Q(2\lambda - 1, 3\lambda + 3, -\lambda - 2) \). The vector \( \overrightarrow{PQ} \) from point \( P \) to point \( Q \) is:
\[
\overrightarrow{PQ} = (2\lambda - 1 - 2, 3\lambda + 3 + 3, -\lambda - 2 - 1)
\]
This simplifies to:
\[
\overrightarrow{PQ} = (2\lambda - 3, 3\lambda + 6, -\lambda - 3)
\]
### Step 4: Set Up the Perpendicular Condition
For \( \overrightarrow{PQ} \) to be perpendicular to the direction ratios of the line, the dot product must equal zero:
\[
(2\lambda - 3, 3\lambda + 6, -\lambda - 3) \cdot (2, 3, -1) = 0
\]
Calculating the dot product:
\[
(2\lambda - 3) \cdot 2 + (3\lambda + 6) \cdot 3 + (-\lambda - 3) \cdot (-1) = 0
\]
This expands to:
\[
4\lambda - 6 + 9\lambda + 18 + \lambda + 3 = 0
\]
Combining like terms gives:
\[
14\lambda + 15 = 0
\]
Thus, solving for \( \lambda \):
\[
\lambda = -\frac{15}{14}
\]
### Step 5: Find the Coordinates of Point Q
Substituting \( \lambda \) back into the parametric equations:
\[
x = 2\left(-\frac{15}{14}\right) - 1 = -\frac{30}{14} - \frac{14}{14} = -\frac{44}{14}
\]
\[
y = 3\left(-\frac{15}{14}\right) + 3 = -\frac{45}{14} + \frac{42}{14} = -\frac{3}{14}
\]
\[
z = -\left(-\frac{15}{14}\right) - 2 = \frac{15}{14} - \frac{28}{14} = -\frac{13}{14}
\]
Thus, the coordinates of point \( Q \) are:
\[
Q\left(-\frac{44}{14}, -\frac{3}{14}, -\frac{13}{14}\right)
\]
### Step 6: Calculate the Distance \( PQ \)
Using the distance formula:
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}
\]
where \( P(2, -3, 1) \) and \( Q\left(-\frac{44}{14}, -\frac{3}{14}, -\frac{13}{14}\right) \):
\[
d = \sqrt{\left(2 + \frac{44}{14}\right)^2 + \left(-3 + \frac{3}{14}\right)^2 + \left(1 + \frac{13}{14}\right)^2}
\]
Calculating each term:
1. \( 2 + \frac{44}{14} = \frac{28}{14} + \frac{44}{14} = \frac{72}{14} \)
2. \( -3 + \frac{3}{14} = -\frac{42}{14} + \frac{3}{14} = -\frac{39}{14} \)
3. \( 1 + \frac{13}{14} = \frac{14}{14} + \frac{13}{14} = \frac{27}{14} \)
Thus, the distance becomes:
\[
d = \sqrt{\left(\frac{72}{14}\right)^2 + \left(-\frac{39}{14}\right)^2 + \left(\frac{27}{14}\right)^2}
\]
Factoring out \( \frac{1}{14^2} \):
\[
d = \frac{1}{14} \sqrt{72^2 + 39^2 + 27^2}
\]
Calculating:
\[
72^2 = 5184, \quad 39^2 = 1521, \quad 27^2 = 729
\]
Adding these:
\[
5184 + 1521 + 729 = 7434
\]
Thus:
\[
d = \frac{\sqrt{7434}}{14}
\]
### Step 7: Final Calculation
Calculating \( \sqrt{7434} \) gives approximately \( 86.22 \):
\[
d \approx \frac{86.22}{14} \approx 6.15
\]
### Final Answer
The length of the perpendicular from point \( P(2, -3, 1) \) to the line is approximately \( 6.15 \) units.
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