Let L _ 1 : → r = ( ^ i - ^ j + 2 ^ k ) + λ( ^ i - ^ j + 2 ^ k ), λ ∈ R , L _ 2 : → r = ( ^ j - ^ k ) + μ(3 ^ i + ^ j + p ^ k ), μ ∈ R and L _ 3 : → r = ( l ^ i + m ^ j + n ^ k ) , ∈ R be three lines such that L _ 1 is

Correct answer: A. ( - 1, 7, 4 ) Given: L _ 1 ⊥ L _ 2 ⇒ ( ^ i - ^ j + 2 ^ k ) . (3 ^ i + ^ j + p ^ k ) = 0 ⇒ 3 - 1 + 2p = 0 ⇒ p = - 1 Also, L _ 3 ⊥ L _ 1 , L _ 2 So, L _ 3 ∥ ( L _ 1 × L _ 2 ) ⇒ L _ 1 × L _ 2 = | cases ^ i & ^ j & ^ k 1 & - 1 & 2 3 & 1 & - 1 cases | ⇒ L _ 1 × L _ 2 = - ^ i + 7 ^ j + 4 ^ k On comparing with L _ 3 : → r = ( l ^ i + m ^ j + n ^ k ) , we get that ( - , 7 , 4 ) will lie on L _ 3 . Now, for = 1 the point will be ( - 1, 7, 4 ) .

JEE Main 202430 JanuaryEvening Shift

Exam: JEE Main
Year: 2024
Date: 30 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Three Dimensional Geometry

JEE Main 2024 · 30 January · Evening Shift · Mathematics · Three Dimensional Geometry

Let L _ 1 : → r = ( ^ i - ^ j + 2 ^ k ) + λ( ^ i - ^ j + 2 ^ k ), λ...

Let

{L}_{1}: \overset{\to}{r} = \left(\overset{^}{i} - \overset{^}{j} + 2\overset{^}{k}\right) + \lambda\left(\overset{^}{i} - \overset{^}{j} + 2\overset{^}{k}\right), \lambda \in R

{L}_{2}: \overset{\to}{r} = \left(\overset{^}{j} - \overset{^}{k}\right) + \mu\left(3\overset{^}{i} + \overset{^}{j} + p\overset{^}{k}\right), \mu \in R

and

{L}_{3}: \overset{\to}{r} = \delta ( l\overset{^}{i} + m\overset{^}{j} + n\overset{^}{k} ) , \delta \in R

be three lines such that

{L}_{1}

is perpendicular to

{L}_{2}

and

{L}_{3}

is perpendicular to both

{L}_{1}

and

{L}_{2}

. Then the point which lies on

{L}_{3}

Options

A $( - 1, 7, 4 )$
B $( - 1, - 7, 4 )$
C $( 1, 7, - 4 )$
D $( 1, - 7, 4 )$

Correct answer

A. $( - 1, 7, 4 )$

Step-by-step solution

Given: ${L}_{1} ⊥ {L}_{2}$
$⇒ \left(\overset{^}{i} - \overset{^}{j} + 2\overset{^}{k}\right) . \left(3\overset{^}{i} + \overset{^}{j} + p\overset{^}{k}\right) = 0$
$⇒ 3 - 1 + 2p = 0$
$⇒ p = - 1$
Also, ${L}_{3} ⊥ {L}_{1}, {L}_{2}$
So, ${L}_{3} ∥ \left({L}_{1} \times {L}_{2}\right)$
$⇒ {L}_{1} \times {L}_{2} = \left|$ \begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ 1 & - 1 & 2 \\ 3 & 1 & - 1\end{cases} $\right|$
$⇒ {L}_{1} \times {L}_{2} = - \overset{^}{i} + 7\overset{^}{j} + 4\overset{^}{k}$
On comparing with ${L}_{3}: \overset{\to}{r} = \delta ( l\overset{^}{i} + m\overset{^}{j} + n\overset{^}{k} )$ , we get that $( - \delta, 7\delta, 4\delta )$ will lie on ${L}_{3}$ .
Now, for $\delta = 1$ the point will be $( - 1, 7, 4 )$ .

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