The shortest distance between the lines x + 1 = 2   y = - 12 z and x = y + 2 = 6 z - 6 is

The shortest distance between the lines x+1=2 y=-12z and x=y+2=6z-6 is
  1. $2$
  2. 3
  3. $\frac{5}{2}$
  4. $\frac{3}{2}$

Solution

Given,

Equation of the lines

x+1=2 y=-12z and x=y+2=6z-6

⇒x+11=y12=z-112 and x1=y+21=z-116

We know that, shortest distance between the lines is given by S.D=b→-a→·p→×q→p→×q→

$p = i + \frac{j}{2} - \frac{k}{12}$ and $q = i + j + \frac{k}{6}$ and $a = i$ and $b = 2j - k$ $\Rightarrow S.D = \frac{-i + 2j - k}{|p \times q|}$ Now solving $p \times q = \begin{vmatrix} i & j & k \\ 1 & \frac{1}{2} & -\frac{1}{12} \\ 1 & 1 & \frac{1}{6} \end{vmatrix}$ $\Rightarrow p \times q = \frac{1}{6}i - \frac{1}{4}j + \frac{1}{2}k$ $\Rightarrow p \times q = 2i - 3j + 6k$ Hence, $S.D = \frac{-i + 2j - k \cdot 2i - 3j + 6k}{\sqrt{2^2 + 3^2 + 6^2}} = |-2| = 2$

Asked in: JEE Main 2023 (25 Jan Shift 2)

Practice more Line and Plane questions on Aicharya