Show that the two line`(x-1)/(2)=(y-2)/(3)=(z-3)/(4) and (x-4)/(5)=(y-1)/(2)=z` intersect. Find also the point of intersection of these lines.

Answer

To show that the two lines intersect and find the point of intersection, we will follow these steps:

### Step 1: Write the parametric equations of the lines
The first line \( L_1 \) is given by:
\[ \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4} \]
Let \( \lambda \) be the parameter. Then we can express the coordinates of points on line \( L_1 \) as:
\[ x = 2\lambda + 1, \quad y = 3\lambda + 2, \quad z = 4\lambda + 3 \]

The second line \( L_2 \) is given by:
\[ \frac{x-4}{5} = \frac{y-1}{2} = z \]
Let \( \mu \) be the parameter. Then we can express the coordinates of points on line \( L_2 \) as:
\[ x = 5\mu + 4, \quad y = 2\mu + 1, \quad z = \mu \]

### Step 2: Set the equations equal to each other
For the lines to intersect, the coordinates must be equal for some values of \( \lambda \) and \( \mu \). Therefore, we set up the following equations:
1. \( 2\lambda + 1 = 5\mu + 4 \) (Equation 1)
2. \( 3\lambda + 2 = 2\mu + 1 \) (Equation 2)
3. \( 4\lambda + 3 = \mu \) (Equation 3)

### Step 3: Solve the equations
From Equation 3, we can express \( \mu \) in terms of \( \lambda \):
\[ \mu = 4\lambda + 3 \]

Now, substitute \( \mu \) into Equations 1 and 2.

**Substituting into Equation 1:**
\[ 2\lambda + 1 = 5(4\lambda + 3) + 4 \]
\[ 2\lambda + 1 = 20\lambda + 15 + 4 \]
\[ 2\lambda + 1 = 20\lambda + 19 \]
Rearranging gives:
\[ 2\lambda - 20\lambda = 19 - 1 \]
\[ -18\lambda = 18 \implies \lambda = -1 \]

**Substituting into Equation 2:**
Now substitute \( \lambda = -1 \) into the expression for \( \mu \):
\[ \mu = 4(-1) + 3 = -4 + 3 = -1 \]

### Step 4: Verify the values of \( \lambda \) and \( \mu \)
Now, we need to check if these values satisfy all three equations.

**Check Equation 1:**
\[ 2(-1) + 1 = 5(-1) + 4 \]
\[ -2 + 1 = -5 + 4 \implies -1 = -1 \quad \text{(True)} \]

**Check Equation 2:**
\[ 3(-1) + 2 = 2(-1) + 1 \]
\[ -3 + 2 = -2 + 1 \implies -1 = -1 \quad \text{(True)} \]

**Check Equation 3:**
\[ 4(-1) + 3 = -1 \]
\[ -4 + 3 = -1 \implies -1 = -1 \quad \text{(True)} \]

### Step 5: Find the point of intersection
Now we can find the point of intersection by substituting \( \lambda = -1 \) into the parametric equations of line \( L_1 \):
\[ x = 2(-1) + 1 = -2 + 1 = -1 \]
\[ y = 3(-1) + 2 = -3 + 2 = -1 \]
\[ z = 4(-1) + 3 = -4 + 3 = -1 \]

Thus, the point of intersection is:
\[ (-1, -1, -1) \]

### Conclusion
The two lines intersect at the point \( (-1, -1, -1) \).

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