QJEEM21Mad567: JEE Main 2021 Mathematics - Class 12 JEEM21Mad567 Solution: If the shortest distance between the straight lines and is ,…

Question

Verified Solution • 2025–26 Session moderate

If the shortest distance between the straight lines

3(x - 1) = 6(y - 2) = 2(z - 1)

and

4(x - 2) = 2(y - \lambda ) = (z - 3),\lambda \in R

is

{1 \over {\sqrt {38} }}

, then the integral value of

\lambda

is equal to :

Accepted Answer

{L_1}:{{(x - 1)} \over 2} = {{(y - 2)} \over 1} = {{(z - 1)} \over 3}\overrightarrow {{r_1}} = 2\widehat i + \widehat j + 3\widehat k

{L_2}:{{(x - 2)} \over 1} = {{y - \lambda } \over 2} = {{z - 3} \over 4}\overrightarrow {{r_2}} = \widehat i + 2\widehat j + 4\widehat k

Shortest distance = Projection of

{\overrightarrow a }

on

{\overrightarrow {{r_1}} \times \overrightarrow {{r_2}} }

= {{\left| {\overrightarrow a .\left( {\overrightarrow {{r_1}} \times \overrightarrow {{r_2}} } \right)} \right|} \over {\left| {\overrightarrow {{r_1}} \times \overrightarrow {{r_2}} } \right|}}

| \vec{a} . (\vec{r_{1}} \times \vec{r_{2}}) | = | \begin{matrix} 1 & λ - 2 & 2 \\ 2 & 1 & 3 \\ 1 & 2 & 4 \end{matrix} | = | 14 - 5 λ |

\left| {\overrightarrow {{r_1}} \times \overrightarrow {{r_2}} } \right| = \sqrt {38}

\therefore

{1 \over {\sqrt {38} }} = {{\left| {14 - 5\lambda } \right|} \over {\sqrt {38} }}

\Rightarrow \left| {14 - 5\lambda } \right| = 1

\Rightarrow 14 - 5\lambda = 1

or

14 - 5\lambda = - 1

\Rightarrow \lambda = {{13} \over 5}

or 3

\therefore

Integral value of

\lambda

= 3.

Step-by-Step Verified Solution