Question

Show that the following lines intersect. Also, find their point of intersection:

Line 1: \[ \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} \]

Line 2: \[ \frac{x - 4}{5} = \frac{y - 1}{2} = z \]

Answer 1

Intersection of Two Lines in 3D

Step 1: Represent the lines in vector form

Line 1 (from symmetric form): \[ \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} = \lambda \Rightarrow \vec{r}_1 = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} + \lambda \begin{bmatrix} 2 \\ 3 \\ 4 \end{bmatrix} \]

Line 2 (from symmetric form): \[ \frac{x - 4}{5} = \frac{y - 1}{2} = z = \mu \Rightarrow \vec{r}_2 = \begin{bmatrix} 4 \\ 1 \\ 0 \end{bmatrix} + \mu \begin{bmatrix} 5 \\ 2 \\ 1 \end{bmatrix} \]

Step 2: Assume the lines intersect

If the lines intersect, then:

\[ \vec{r}_1 = \vec{r}_2 \Rightarrow \begin{bmatrix} 1 + 2\lambda \\ 2 + 3\lambda \\ 3 + 4\lambda \end{bmatrix} = \begin{bmatrix} 4 + 5\mu \\ 1 + 2\mu \\ \mu \end{bmatrix} \]

Equating components:

• (i) \( 1 + 2\lambda = 4 + 5\mu \)
• (ii) \( 2 + 3\lambda = 1 + 2\mu \)
• (iii) \( 3 + 4\lambda = \mu \)

Step 3: Solve the system

From (iii): \[ \mu = 3 + 4\lambda \] Substitute into (i): \[ 1 + 2\lambda = 4 + 5(3 + 4\lambda) = 4 + 15 + 20\lambda = 19 + 20\lambda \Rightarrow 1 + 2\lambda = 19 + 20\lambda \Rightarrow -18 = 18\lambda \Rightarrow \lambda = -1 \]

Now, from (iii): \[ \mu = 3 + 4(-1) = 3 - 4 = -1 \]

Check in (ii): \[ 2 + 3(-1) = -1,\quad 1 + 2(-1) = -1 \Rightarrow \text{LHS = RHS ✅}

Step 4: Find the point of intersection

Substitute \( \lambda = -1 \) into Line 1:

\[ \vec{r} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} + (-1) \cdot \begin{bmatrix} 2 \\ 3 \\ 4 \end{bmatrix} = \begin{bmatrix} 1 - 2 \\ 2 - 3 \\ 3 - 4 \end{bmatrix} = \begin{bmatrix} -1 \\ -1 \\ -1 \end{bmatrix} \]

✅ Final Answer:

The lines intersect at the point: \[ \boxed{(-1,\ -1,\ -1)} \]