If the lines $\frac{x - 5}{5\lambda + 2} = \frac{2 - y}{5} = \frac{1 - z}{-1}$ and $x = \frac{y + 1/2}{2\lambda} = \frac{z - 1}{3}$ are perpendicular, then the value of $\lambda$ is equal to

We have two lines in 3D space, and we need to find the value of $\lambda$ that makes them perpendicular.

Given Line 1:

$\dfrac{x - 5}{5\lambda + 2} = \dfrac{2 - y}{5} = \dfrac{1 - z}{-1}$

We need to rewrite this in standard form, where the numerator must be (variable minus constant), not the other way around.

$\dfrac{2 - y}{5} = \dfrac{-(y - 2)}{5} = \dfrac{y - 2}{-5}$

$\dfrac{1 - z}{-1} = \dfrac{-(z - 1)}{-1} = \dfrac{z - 1}{1}$

So Line 1 becomes:

$\dfrac{x - 5}{5\lambda + 2} = \dfrac{y - 2}{-5} = \dfrac{z - 1}{1}$

Direction ratios of Line 1: $(5\lambda + 2,\ -5,\ 1)$

Given Line 2:

$x = \dfrac{y + \frac{1}{2}}{2\lambda} = \dfrac{z - 1}{3}$

Here, $x$ appears alone, which is the same as writing $\dfrac{x - 0}{1}$, so:

$\dfrac{x - 0}{1} = \dfrac{y - (-\frac{1}{2})}{2\lambda} = \dfrac{z - 1}{3}$

Direction ratios of Line 2: $(1,\ 2\lambda,\ 3)$

Two lines are perpendicular when the dot product of their direction ratios equals zero, i.e. $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$

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

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

$-5\lambda + 5 = 0$

$\lambda = 1$

Therefore, $\lambda = 1$