To solve the problem, we need to find the shortest distance between two lines given certain conditions. Here’s a step-by-step breakdown of the solution:
### Step 1: Identify the Points and the Line
We have the point \( P(0, 3, 1) \) from which the perpendicular is drawn to the line given by the equation:
\[
\frac{x-a}{l} = \frac{y-2}{3} = \frac{z-b}{4}
\]
The foot of the perpendicular is the point \( Q(1, -1, 2) \).
### Step 2: Find Direction Ratios
The direction ratios of the line can be denoted as \( (l, 3, 4) \). The direction ratios of the line segment \( PQ \) (from \( P \) to \( Q \)) can be calculated as:
\[
PQ = (1 - 0, -1 - 3, 2 - 1) = (1, -4, 1)
\]
### Step 3: Use the Perpendicular Condition
For the lines to be perpendicular, the dot product of their direction ratios must equal zero:
\[
(l, 3, 4) \cdot (1, -4, 1) = 0
\]
Calculating the dot product:
\[
l \cdot 1 + 3 \cdot (-4) + 4 \cdot 1 = 0
\]
This simplifies to:
\[
l - 12 + 4 = 0 \implies l - 8 = 0 \implies l = 8
\]
### Step 4: Find Values of \( a \) and \( b \)
Since point \( Q(1, -1, 2) \) lies on the line, we substitute \( x = 1 \), \( y = -1 \), and \( z = 2 \) into the line equation:
\[
\frac{1-a}{8} = \frac{-1-2}{3} = \frac{2-b}{4}
\]
Calculating each part:
1. From \( \frac{1-a}{8} = \frac{-3}{3} \):
\[
\frac{1-a}{8} = -1 \implies 1 - a = -8 \implies a = 9
\]
2. From \( \frac{2-b}{4} = -1 \):
\[
2 - b = -4 \implies b = 6
\]
### Step 5: Write the Line Equations
Now we have:
- The first line:
\[
\frac{x-9}{8} = \frac{y-2}{3} = \frac{z-6}{4}
\]
- The second line:
\[
\frac{x-1}{3} = \frac{y-2}{4} = \frac{z-3}{5}
\]
### Step 6: Determine Direction Ratios and Points
For the first line \( L_1 \):
- Point: \( (9, 2, 6) \)
- Direction ratios: \( (8, 3, 4) \)
For the second line \( L_2 \):
- Point: \( (1, 2, 3) \)
- Direction ratios: \( (3, 4, 5) \)
### Step 7: Use the Shortest Distance Formula
The formula for the shortest distance \( d \) between two skew lines is given by:
\[
d = \frac{|(P_2 - P_1) \cdot (D_1 \times D_2)|}{|D_1 \times D_2|}
\]
Where \( P_1 \) and \( P_2 \) are points on the lines, and \( D_1 \) and \( D_2 \) are the direction ratios.
Calculating \( P_2 - P_1 \):
\[
(1 - 9, 2 - 2, 3 - 6) = (-8, 0, -3)
\]
Calculating the cross product \( D_1 \times D_2 \):
\[
D_1 = (8, 3, 4), \quad D_2 = (3, 4, 5)
\]
Using the determinant:
\[
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
8 & 3 & 4 \\
3 & 4 & 5
\end{vmatrix}
= \hat{i}(3 \cdot 5 - 4 \cdot 4) - \hat{j}(8 \cdot 5 - 4 \cdot 3) + \hat{k}(8 \cdot 4 - 3 \cdot 3)
\]
Calculating:
\[
= \hat{i}(15 - 16) - \hat{j}(40 - 12) + \hat{k}(32 - 9)
= \hat{i}(-1) - \hat{j}(28) + \hat{k}(23)
= (-1, -28, 23)
\]
### Step 8: Calculate the Magnitude of the Cross Product
\[
|D_1 \times D_2| = \sqrt{(-1)^2 + (-28)^2 + (23)^2} = \sqrt{1 + 784 + 529} = \sqrt{1314}
\]
### Step 9: Calculate the Dot Product
\[
|(P_2 - P_1) \cdot (D_1 \times D_2)| = |(-8, 0, -3) \cdot (-1, -28, 23)| = |-8 \cdot (-1) + 0 \cdot (-28) + (-3) \cdot 23| = |8 - 69| = | -61| = 61
\]
### Step 10: Final Distance Calculation
\[
d = \frac{61}{\sqrt{1314}}
\]
