Question

Represent the equations of lines $l_1$ and $l_2$ in vector form and check whether they are intersecting or not.
$l_1 : \frac{x+3}{-3} = \frac{y-1}{1} = \frac{z-5}{5}$
$l_2 : \frac{x+1}{-1} = \frac{2-y}{-2} = \frac{z-5}{5}$}

Hint:

Always standardize the equations first (check the coefficient of $y$ in $l_2$). If the three equations for $\lambda$ and $\mu$ are inconsistent, the lines are skew; if they have a common solution, they intersect.

Answer 1

Concept:
• Vector form: $\vec{r} = \vec{a} + \lambda \vec{b}$, where $\vec{a}$ is a point on the line and $\vec{b}$ is the direction vector.
• Lines intersect if there exist $\lambda, \mu$ such that $\vec{a_1} + \lambda \vec{b_1} = \vec{a_2} + \mu \vec{b_2}$.

Step 1: Write lines in standard vector form
For $l_1$: point is $(-3, 1, 5)$, direction is $\langle -3, 1, 5 \rangle$.
$\vec{r} = (-3\hat{i} + \hat{j} + 5\hat{k}) + \lambda(-3\hat{i} + \hat{j} + 5\hat{k})$.
For $l_2$: point is $(-1, 2, 5)$, direction is $\langle -1, 2, 5 \rangle$ (note $\frac{y-2}{2}$ conversion).
$\vec{r} = (-\hat{i} + 2\hat{j} + 5\hat{k}) + \mu(-\hat{i} + 2\hat{j} + 5\hat{k})$.

Step 2: Set up the intersection equations
$-3 - 3\lambda = -1 - \mu \implies \mu - 3\lambda = 2 \quad \dots(1)$
$1 + \lambda = 2 + 2\mu \implies \lambda - 2\mu = 1 \quad \dots(2)$
$5 + 5\lambda = 5 + 5\mu \implies \lambda = \mu \quad \dots(3)$

Step 3: Solve the system
From (3), substitute $\mu = \lambda$ into (1):
$\lambda - 3\lambda = 2 \implies -2\lambda = 2 \implies \lambda = -1, \mu = -1$.
Check in (2): $-1 - 2(-1) = -1 + 2 = 1$. (True)
Since a common pair exists, the lines intersect.

Step 4: Find the intersection point
Substitute $\lambda = -1$ into $l_1$:
$x = -3 - 3(-1) = 0$
$y = 1 + (-1) = 0$
$z = 5 + 5(-1) = 0$
The lines intersect at the origin $(0, 0, 0)$.