2. With respect to a fixed origin [tex]\( O \)[/tex], the lines [tex]\( l_1 \)[/tex] and [tex]\( l_2 \)[/tex] are given by the equations:

[tex]\[
\begin{array}{l}
l_1: \mathbf{r} = (-3\mathbf{i} + 5\mathbf{k}) + \lambda(5\mathbf{i} - \mathbf{j} + \mathbf{k}) \\
l_2: \mathbf{r} = (10\mathbf{i} - \mathbf{j} + 15\mathbf{k}) + \mu(6\mathbf{i} - 2\mathbf{j} + 4\mathbf{k})
\end{array}
\][/tex]

Show that lines [tex]\( l_1 \)[/tex] and [tex]\( l_2 \)[/tex] do not meet.



Answer :

To show that the lines [tex]\( l_1 \)[/tex] and [tex]\( l_2 \)[/tex] do not meet, we need to investigate their positions and directions in space and determine if they intersect. Given the equations:

[tex]\[ \begin{array}{l} l_1: \vec{r}_1 = (-3\hat{i} + 5\hat{k}) + \lambda(5\hat{i} - \hat{j} + \hat{k}) \\ l_2: \vec{r}_2 = (10\hat{i} - \hat{j} + 15\hat{k}) + \mu(6\hat{i} - 2\hat{j} + 4\hat{k}) \end{array} \][/tex]

where [tex]\( \lambda \)[/tex] and [tex]\( \mu \)[/tex] are parameters. Let's break this down step-by-step to check if these lines intersect.

Step 1: Identify points and direction vectors

- The point on line [tex]\( l_1 \)[/tex] can be written as [tex]\( \vec{A} = -3\hat{i} + 5\hat{k} \)[/tex].
- The direction vector of [tex]\( l_1 \)[/tex] is [tex]\( \vec{d}_1 = 5\hat{i} - \hat{j} + \hat{k} \)[/tex].

- The point on line [tex]\( l_2 \)[/tex] can be written as [tex]\( \vec{B} = 10\hat{i} - \hat{j} + 15\hat{k} \)[/tex].
- The direction vector of [tex]\( l_2 \)[/tex] is [tex]\( \vec{d}_2 = 6\hat{i} - 2\hat{j} + 4\hat{k} \)[/tex].

Step 2: Form the system to solve for intersection

For the lines [tex]\( l_1 \)[/tex] and [tex]\( l_2 \)[/tex] to intersect, there must exist values of [tex]\( \lambda \)[/tex] and [tex]\( \mu \)[/tex] such that:
[tex]\[ \vec{A} + \lambda \vec{d}_1 = \vec{B} + \mu \vec{d}_2 \][/tex]

This results in the following system of equations:
[tex]\[ \begin{cases} -3 + 5\lambda = 10 + 6\mu \\ 0 - \lambda = -1 - 2\mu \\ 5 + \lambda = 15 + 4\mu \end{cases} \][/tex]

Step 3: Transform into matrix form and analyze ranks

Rewriting this system in matrix form for clarity:

[tex]\[ \begin{pmatrix} 5 & -6 & 13 \\ -1 & 2 & -1 \\ 1 & -4 & 10 \end{pmatrix} \begin{pmatrix} \lambda \\ \mu \\ \end{pmatrix} = \begin{pmatrix} 13 \\ -1 \\ 10 \end{pmatrix} \][/tex]

Here, the matrix consists of 2 matrices:
- The coefficient matrix:
[tex]\[ C = \begin{pmatrix} 5 & -6 \\ -1 & 2 \\ 1 & -4 \end{pmatrix} \][/tex]

- The augmented matrix:
[tex]\[ A = \begin{pmatrix} 5 & -6 & 13 \\ -1 & 2 & -1 \\ 1 & -4 & 10 \end{pmatrix} \][/tex]

Step 4: Determine ranks

- The rank of the coefficient matrix [tex]\( C \)[/tex] is 2.
- The rank of the augmented matrix [tex]\( A \)[/tex] is 3.

Step 5: Interpret ranks

The rank of the coefficient matrix (2) is less than the rank of the augmented matrix (3). According to the Rouché–Capelli theorem, since the ranks are different, the system of equations is inconsistent.

Therefore, the lines [tex]\( l_1 \)[/tex] and [tex]\( l_2 \)[/tex] do not meet, confirming they do not intersect in 3D space.