Question

Show that the lines
\[ \frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7} \quad \text{and} \quad \frac{x - 2}{1} = \frac{y - 4}{3} = \frac{z - 6}{5} \] intersect and find their point of intersection.

Answer 1

Solution:
Step 1: Write the equations in parametric form.
First line:
\[ x = -1 + 3s,\quad y = -3 + 5s,\quad z = -5 + 7s \]
Second line:
\[ x = 2 + t,\quad y = 4 + 3t,\quad z = 6 + 5t \]
Step 2: At intersection, equate the coordinates:
\[ -1 + 3s = 2 + t \Rightarrow t = 3s - 3 \]
Substitute in y equation: \[ -3 + 5s = 4 + 3(3s - 3) \Rightarrow -3 + 5s = 4 + 9s - 9 \Rightarrow -3 + 5s = -5 + 9s \Rightarrow 4s = 2 \Rightarrow s = \frac{1}{2} \]
Then, \[ t = 3\left(\frac{1}{2}\right) - 3 = \frac{3}{2} - 3 = -\frac{3}{2} \]
Check in z equation: \[ -5 + 7\left(\frac{1}{2}\right) = 6 + 5\left(-\frac{3}{2}\right) \Rightarrow -5 + 3.5 = -1.5,\quad 6 - 7.5 = -1.5 \]
LHS = RHS, so valid.
Step 3: Find point of intersection using \( s = \frac{1}{2} \):
\[ x = -1 + 3\left(\frac{1}{2}\right) = \frac{1}{2},\quad y = -3 + 5\left(\frac{1}{2}\right) = -\frac{1}{2},\quad z = -5 + 7\left(\frac{1}{2}\right) = -\frac{3}{2} \]
Final Answer: The lines intersect at the point \[ \left(\frac{1}{2}, -\frac{1}{2}, -\frac{3}{2}\right) \]