If the shortest distance between the lines x – 2 = y – 2 1 = z – 1 1 and x – 3 1 = y – 1 – 2 = z – 2 1 is 1, then the sum of all possible values of is: [2024]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
If the shortest distance between the lines $\frac{x - λ}{- 2} = \frac{y - 2}{1} = \frac{z - 1}{1}$ and $\frac{x - \sqrt{3}}{1} = \frac{y - 1}{- 2} = \frac{z - 2}{1}$ is 1, then the sum of all possible values of $λ$ is: [2024]

1 $2 \sqrt{3}$

2 0

3 $3 \sqrt{3}$

4 $- 2 \sqrt{3}$

Ans.
(1)

The shortest distance between the given lines is given by

$d =$ $| \frac{| \begin{matrix} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{matrix} |}{\sqrt{{(a_{1} b_{2} - a_{2} b_{1})}^{2} + {(b_{1} c_{2} - b_{2} c_{1})}^{2} + {(c_{1} a_{2} - c_{2} a_{1})}^{2}}} |$

$\Rightarrow 1 =$ $| \frac{| \begin{matrix} \sqrt{3} - λ & - 1 & 1 \\ - 2 & 1 & 1 \\ 1 & - 2 & 1 \end{matrix} |}{\sqrt{{(3)}^{2} + {(3)}^{2} + {(3)}^{2}}} |$

$\Rightarrow 1 = \frac{| (\sqrt{3} - λ) (3) + 1 (- 3) + 1 (3) |}{\sqrt{27}}$

$\Rightarrow 3 \sqrt{3} = 3 \sqrt{3} - 3 λ or - 3 \sqrt{3} = 3 \sqrt{3} - 3 λ$

$\Rightarrow λ = 0 or 6 \sqrt{3} = 3 λ \Rightarrow λ = 2 \sqrt{3}$

$∴$ The sum of possible values of $λ = 0 + 2 \sqrt{3} = 2 \sqrt{3}$ .

Want Us to Email you About Special Offers & Updates?