Q.Find the value of $\lambda$ for which the following lines are perpendicular to each other: $\dfrac{x-5}{5\lambda+2} = \dfrac{2-y}{5} = \dfrac{1-z}{-1}$; $\dfrac{x}{1} = \dfrac{y + \frac{1}{2}}{2\lambda} = \dfrac{z-1}{3}$; hence, find whether the lines intersect or not.

Concept: Two lines are perpendicular if the dot product of their direction vectors is zero.

Step 1 – Direction vectors

First line: $\frac{x-5}{5\lambda+2} = \frac{y-2}{-5} = \frac{z-1}{1}$ (rewriting $2-y$ as $-(y-2)$ and $1-z$ as $-(z-1)$).

Direction vector: $\vec{d}_1 = (5\lambda+2,\ -5,\ 1)$.

Second line: $\frac{x}{1} = \frac{y+\frac12}{2\lambda} = \frac{z-1}{3}$.

Direction vector: $\vec{d}_2 = (1,\ 2\lambda,\ 3)$.

Step 2 – Perpendicular condition

$\vec{d}_1 \cdot \vec{d}_2 = 0$

$(5\lambda+2)(1) + (-5)(2\lambda) + (1)(3) = 0$

$5\lambda + 2 - 10\lambda + 3 = 0$

$-5\lambda + 5 = 0 \implies \lambda = 1$.

Step 3 – Check intersection

For $\lambda=1$, lines are:

$L_1: \frac{x-5}{7} = \frac{y-2}{-5} = \frac{z-1}{1}$, point $A(5,2,1)$.

$L_2: \frac{x}{1} = \frac{y+\frac12}{2} = \frac{z-1}{3}$, point $B(0,-\frac12,1)$.

Parametric forms:

$L_1: (5+7t,\ 2-5t,\ 1+t)$

$L_2: (s,\ -\frac12+2s,\ 1+3s)$ …

Why direction vectors decide perpendicularity

Two lines in space are perpendicular when their direction vectors are perpendicular — that is, their dot product equals zero. The direction vector of a line given in symmetric form $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$ is simply $(a, b, c)$. So the entire problem reduces to reading off these numbers correctly and solving a simple equation.

But there's a subtlety: the second line's equation has $2-y$ in the denominator, not $y - 2$. That flips the sign of that component. Let's handle that carefully.

1. Extract the direction vector of the first line

   The first line is:

Rewrite $\frac{2-y}{5}$ as $\frac{y-2}{-5}$ — because $2-y = -(y-2)$. Similarly, $\frac{1-z}{-1} = \frac{z-1}{1}$.

So the symmetric form becomes:

Hence the direction vector of the first line is:

2. Extract the direction vector of the second line

   The second line is:

This is already in standard form. Its direction vector is:

3. Apply the perpendicular condition

   For perpendicular lines:

Compute:

4. Now check whether the lines intersect

   With $\lambda = 1$, the lines become:

   Line 1: $\frac{x-5}{7} = \frac{y-2}{-5} = \frac{z-1}{1} = t$ (say)

   So parametric form:

Line 2: $\frac{x}{1} = \frac{y + \frac12}{2} = \frac{z-1}{3} = s$ (say) …