The shortest distance between lines L _ 1 and L _ 2 , where L _ 1 : (x - 1)/(2) = (y + 1)/(- 3) = (z + 4)/(2) and L _ 2 is the line passing through the points A (- 4, 4, 3), B(- 1, 6, 3) and perpendicular to the line (x

Correct answer: C. 141 221 Given: L _ 2 passes through A(- 4, 4, 3)andB(- 1, 6, 3) . ⇒ L _ 2 : (x + 4)/(3) = (y - 4)/(2) = (z - 3)/(0) Also, L _ 1 : (x - 1)/(2) = (y + 1)/(- 3) = (z + 4)/(2) We know that, shortest distance between L _ 1 : x - x _ 1 a _ 1 = y - y _ 1 b _ 1 = z - z _ 1 c _ 1 and L _ 2 : x - x _ 2 a _ 2 = y - y _ 2 b _ 2 = z - z _ 2 c _ 2 is given by, | | cases x _ 2 - x _ 1 & y _ 2 - y _ 1 & z _ 2 - z _ 1 a _ 1 & b _ 1 & c _ 1 a _ 2 & b _ 2 & c _ 2 cases | ( 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 | . So, shortest distance, D = | | cases - 4 - 1 & 1 + 4 & 4 + 3 2 & - 3 & 2 3 & 2 & 0 cases | (4 + 9) ^ 2 + (0 - 4) ^ 2 + (6 - 0) ^ 2 | ⇒ D = | | cases - 5 & 5 & 7 2 & - 3 & 2 3 & 2 & 0 cases | 169 + 16 + 36 | ⇒ D = | 31 × 3 + 48 221 | ⇒ D = 141 221

JEE Main 202431 JanuaryEvening Shift

Exam: JEE Main
Year: 2024
Date: 31 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Three Dimensional Geometry

JEE Main 2024 · 31 January · Evening Shift · Mathematics · Three Dimensional Geometry

The shortest distance between lines L _ 1 and L _ 2 , where L _ 1 :...

The shortest distance between lines

{L}_{1}

and

{L}_{2}

, where

{L}_{1}: \dfrac{x - 1}{2} = \dfrac{y + 1}{- 3} = \dfrac{z + 4}{2}

and

{L}_{2}

is the line passing through the points

A

\left(- 4, 4, 3\right),

B\left(- 1, 6, 3\right)

and perpendicular to the line

\dfrac{x - 3}{- 2} = \dfrac{y}{3} = \dfrac{z - 1}{1}

, is

Options

A $\dfrac{121}{\sqrt{221}}$
B $\dfrac{24}{\sqrt{117}}$
C $\dfrac{141}{\sqrt{221}}$
D $\dfrac{42}{\sqrt{117}}$

Correct answer

C. $\dfrac{141}{\sqrt{221}}$

Step-by-step solution

Given: ${L}_{2}$ passes through $A\left(- 4, 4, 3\right)\text{and}B\left(- 1, 6, 3\right)$ .

$⇒ {L}_{2}: \dfrac{x + 4}{3} = \dfrac{y - 4}{2} = \dfrac{z - 3}{0}$

Also, ${L}_{1}: \dfrac{x - 1}{2} = \dfrac{y + 1}{- 3} = \dfrac{z + 4}{2}$

We know that, shortest distance between ${L}_{1}: \dfrac{x - {x}_{1}}{{a}_{1}} = \dfrac{y - {y}_{1}}{{b}_{1}} = \dfrac{z - {z}_{1}}{{c}_{1}}$ and ${L}_{2}: \dfrac{x - {x}_{2}}{{a}_{2}} = \dfrac{y - {y}_{2}}{{b}_{2}} = \dfrac{z - {z}_{2}}{{c}_{2}}$ is given by,

\left|\dfrac{\left| $\begin{cases}{x}_{2} - {x}_{1} & {y}_{2} - {y}_{1} & {z}_{2} - {z}_{1} \\ {a}_{1} & {b}_{1} & {c}_{1} \\ {a}_{2} & {b}_{2} & {c}_{2}\end{cases}$ \right|}{\sqrt{{\left({a}_{1}{b}_{2} - {a}_{2}{b}_{1}\right)}^{2} + {\left({b}_{1}{c}_{2} - {b}_{2}{c}_{1}\right)}^{2} + {\left({c}_{1}{a}_{2} - {c}_{2}{a}_{1}\right)}^{2}}}\right| $.So, shortest distance,$ D = \left|\dfrac{\left| $\begin{cases}- 4 - 1 & 1 + 4 & 4 + 3 \\ 2 & - 3 & 2 \\ 3 & 2 & 0\end{cases}$ \right|}{\sqrt{{\left(4 + 9\right)}^{2} + {\left(0 - 4\right)}^{2} + {\left(6 - 0\right)}^{2}}}\right| $$ ⇒ D = \left|\dfrac{\left| $\begin{cases}- 5 & 5 & 7 \\ 2 & - 3 & 2 \\ 3 & 2 & 0\end{cases}$ \right|}{\sqrt{169 + 16 + 36}}\right| $$ ⇒ D = \left|\dfrac{31 \times 3 + 48}{\sqrt{221}}\right| $$ ⇒ D = \dfrac{141}{\sqrt{221}}$

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