To determine the relationship between the two lines given by the equations:
1. \(\frac{x-1}{3} = \frac{y+1}{2} = \frac{z-1}{5}\)
2. \(x = \frac{y-1}{3} = \frac{z+1}{-2}\)
we will follow these steps:
### Step 1: Identify Points and Direction Ratios of the Lines
**Line 1:**
From the equation \(\frac{x-1}{3} = \frac{y+1}{2} = \frac{z-1}{5}\), we can express it in parametric form:
- Let \(t = \frac{x-1}{3}\), then:
- \(x = 3t + 1\)
- \(y = 2t - 1\)
- \(z = 5t + 1\)
Thus, the point on Line 1 is \(P_1(1, -1, 1)\) when \(t = 0\) and the direction ratios are \( (3, 2, 5) \).
**Line 2:**
From the equation \(x = \frac{y-1}{3} = \frac{z+1}{-2}\), we can express it in parametric form:
- Let \(s = x\), then:
- \(y = 3s + 1\)
- \(z = -2s - 1\)
Thus, the point on Line 2 is \(P_2(0, 1, -1)\) when \(s = 0\) and the direction ratios are \( (1, 3, -2) \).
### Step 2: Check for Parallelism
To check if the lines are parallel, we compare the direction ratios:
- For Line 1: \( (3, 2, 5) \)
- For Line 2: \( (1, 3, -2) \)
We check the ratios:
\[
\frac{3}{1} \neq \frac{2}{3} \neq \frac{5}{-2}
\]
Since the ratios are not equal, the lines are not parallel.
### Step 3: Check for Perpendicularity
To check if the lines are perpendicular, we calculate the dot product of the direction ratios:
\[
(3, 2, 5) \cdot (1, 3, -2) = 3 \cdot 1 + 2 \cdot 3 + 5 \cdot (-2) = 3 + 6 - 10 = -1
\]
Since the dot product is not zero, the lines are not perpendicular.
### Step 4: Check for Coplanarity
To check if the lines are coplanar, we use the condition involving the scalar triple product. We need to calculate:
1. \(P_2 - P_1 = (0 - 1, 1 - (-1), -1 - 1) = (-1, 2, -2)\)
2. The direction ratios of the lines:
- \(d_1 = (3, 2, 5)\)
- \(d_2 = (1, 3, -2)\)
Now we form the matrix:
\[
\begin{vmatrix}
-1 & 2 & -2 \\
3 & 2 & 5 \\
1 & 3 & -2
\end{vmatrix}
\]
Calculating the determinant:
\[
= -1 \begin{vmatrix} 2 & 5 \\ 3 & -2 \end{vmatrix} - 2 \begin{vmatrix} 3 & 5 \\ 1 & -2 \end{vmatrix} - 2 \begin{vmatrix} 3 & 2 \\ 1 & 3 \end{vmatrix}
\]
Calculating each of the 2x2 determinants:
1. \(\begin{vmatrix} 2 & 5 \\ 3 & -2 \end{vmatrix} = (2)(-2) - (5)(3) = -4 - 15 = -19\)
2. \(\begin{vmatrix} 3 & 5 \\ 1 & -2 \end{vmatrix} = (3)(-2) - (5)(1) = -6 - 5 = -11\)
3. \(\begin{vmatrix} 3 & 2 \\ 1 & 3 \end{vmatrix} = (3)(3) - (2)(1) = 9 - 2 = 7\)
Now substituting back:
\[
= -1(-19) - 2(-11) - 2(7) = 19 + 22 - 14 = 27
\]
Since the determinant is not zero, the lines are not coplanar.
### Conclusion
Since the lines are neither parallel, nor perpendicular, nor coplanar, we conclude that the lines do not intersect.
