Find the distance of the point (1,-2,3) from the line joining the points (-1,2,5) and (2,3,4).

Answer

To find the distance of the point \( P(1, -2, 3) \) from the line joining the points \( A(-1, 2, 5) \) and \( B(2, 3, 4) \), we follow these steps:

### Step 1: Find the direction ratios of the line AB
The direction ratios of the line joining points \( A \) and \( B \) can be calculated as follows:
\[ \text{Direction ratios} = (B_x - A_x, B_y - A_y, B_z - A_z) = (2 - (-1), 3 - 2, 4 - 5) = (3, 1, -1) \]

### Step 2: Write the parametric equations of the line
Using the point \( A \) and the direction ratios, we can write the parametric equations of the line:
\[ \frac{x + 1}{3} = \frac{y - 2}{1} = \frac{z - 5}{-1} = r \]
From this, we can express \( x, y, z \) in terms of \( r \):
\[ x = 3r - 1, \quad y = r + 2, \quad z = -r + 5 \]

### Step 3: Find the direction ratios of line PC
Let \( C(x, y, z) \) be a point on the line \( AB \). The direction ratios of the line \( PC \) (from point \( P(1, -2, 3) \) to point \( C \)) are:
\[ \text{Direction ratios of } PC = (x - 1, y + 2, z - 3) \]

### Step 4: Set up the perpendicularity condition
Since \( PC \) is perpendicular to \( AB \), we can use the dot product of the direction ratios:
\[ (3, 1, -1) \cdot (x - 1, y + 2, z - 3) = 0 \]
Substituting the values of \( x, y, z \):
\[ 3(3r - 1 - 1) + 1(r + 2 + 2) - 1(-r + 5 - 3) = 0 \]
This simplifies to:
\[ 3(3r - 2) + (r + 4) + (r - 2) = 0 \]
\[ 9r - 6 + r + 4 + r - 2 = 0 \]
\[ 11r - 4 = 0 \]
Thus, solving for \( r \):
\[ r = \frac{4}{11} \]

### Step 5: Find the coordinates of point C
Substituting \( r = \frac{4}{11} \) back into the equations for \( x, y, z \):
\[ x = 3\left(\frac{4}{11}\right) - 1 = \frac{12}{11} - \frac{11}{11} = \frac{1}{11} \]
\[ y = \frac{4}{11} + 2 = \frac{4}{11} + \frac{22}{11} = \frac{26}{11} \]
\[ z = -\frac{4}{11} + 5 = -\frac{4}{11} + \frac{55}{11} = \frac{51}{11} \]
So, the coordinates of point \( C \) are \( \left(\frac{1}{11}, \frac{26}{11}, \frac{51}{11}\right) \).

### Step 6: Calculate the distance PC
Now, we can find the distance \( PC \) using the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]
Substituting \( P(1, -2, 3) \) and \( C\left(\frac{1}{11}, \frac{26}{11}, \frac{51}{11}\right) \):
\[ d = \sqrt{\left(1 - \frac{1}{11}\right)^2 + \left(-2 - \frac{26}{11}\right)^2 + \left(3 - \frac{51}{11}\right)^2} \]
Calculating each term:
1. \( 1 - \frac{1}{11} = \frac{10}{11} \)
2. \( -2 - \frac{26}{11} = -\frac{22}{11} - \frac{26}{11} = -\frac{48}{11} \)
3. \( 3 - \frac{51}{11} = \frac{33}{11} - \frac{51}{11} = -\frac{18}{11} \)

Thus, we have:
\[ d = \sqrt{\left(\frac{10}{11}\right)^2 + \left(-\frac{48}{11}\right)^2 + \left(-\frac{18}{11}\right)^2} \]
Calculating:
\[ d = \sqrt{\frac{100}{121} + \frac{2304}{121} + \frac{324}{121}} = \sqrt{\frac{2728}{121}} = \frac{\sqrt{2728}}{11} \]

### Final Answer
The distance of the point \( (1, -2, 3) \) from the line joining the points \( (-1, 2, 5) \) and \( (2, 3, 4) \) is:
\[ \frac{\sqrt{2728}}{11} \]