Let be the point on the line at a distance from the point and nearer to the origin. Then the shortest distance between the lines , is equal to [2026]

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.
Let $P (α, β, γ)$ be the point on the line $\frac{x - 1}{2} = \frac{y + 1}{- 3} = z$ at a distance $4 \sqrt{14}$ from the point $(1, - 1, 0)$ and nearer to the origin. Then the shortest distance between the lines $\frac{x - α}{1} = \frac{y - β}{2} = \frac{z - γ}{3} and \frac{x + 5}{2} = \frac{y - 10}{1} = \frac{z - 3}{1}$ , is equal to [2026]

1 $7 \sqrt{\frac{5}{4}}$

2 $2 \sqrt{\frac{7}{4}}$

3 $4 \sqrt{\frac{7}{5}}$

4 $4 \sqrt{\frac{5}{7}}$

Ans.
(3)

$Let P (2 λ + 1, - 3 λ - 1, λ)$

$Then 4 λ^{2} + 9 λ^{2} + λ^{2} = 16 \cdot 14 \Rightarrow λ = \pm 4$ $\Rightarrow - 4 (nearer to origin)$

$∴$ $P (- 7, 11, - 4)$

$∴$ $Shortest distance = \frac{| \begin{matrix} 2 & - 1 & 7 \\ 1 & 2 & 3 \\ 2 & 1 & 1 \end{matrix} |}{| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 2 & 1 & 1 \end{matrix} |}$

$= \frac{28}{\sqrt{1 + 25 + 9}} = \frac{4 \sqrt{7}}{\sqrt{5}}$

Want Us to Email you About Special Offers & Updates?