- Exam
- JEE Advanced
- Year
- 2019
- Paper
- Paper 1
- Subject
- Mathematics
- Chapter
- Three Dimensional Geometry
JEE Advanced 2019 · Paper 1 · Mathematics · Three Dimensional Geometry
Let L _ 1 and L _ 2 denotes the lines → r = ^ i + λ(- ^ i + 2 ^ j +...
\overset{\to}{r} = \overset{^}{i} + \lambda\left(- \overset{^}{i} + 2\overset{^}{j} + 2\overset{^}{k}\right), \lambda \in R and \overset{\to}{r} = \mu\left(2\overset{^}{i} - \overset{^}{j} + 2\overset{^}{k}\right), \mu \in R
respectively. If is a line which is perpendicular to both and and cuts both of them, then which of the following options describe(s) ?
Options
- A\overset{\to}{r} = \dfrac{1}{3}\left(2\overset{^}{i} + \overset{^}{k}\right) + t\left(2\overset{^}{i} + 2\overset{^}{j} - \overset{^}{k}\right), t \in R
- B\overset{\to}{r} = \dfrac{2}{9}\left(4\overset{^}{i} + \overset{^}{j} + \overset{^}{k}\right) + t\left(2\overset{^}{i} + 2\overset{^}{j} - \overset{^}{k}\right), t \in R
- C\overset{\to}{r} = \dfrac{2}{9}\left(2\overset{^}{i} - \overset{^}{j} + 2\overset{^}{k}\right), t\left(2\overset{^}{i} + 2\overset{^}{j} - \overset{^}{k}\right), t \in R
- D\overset{\to}{r} = t\left(2\overset{^}{i} + 2\overset{^}{j} - \overset{^}{k}\right), t \in R
Correct answer
A. \overset{\to}{r} = \dfrac{1}{3}\left(2\overset{^}{i} + \overset{^}{k}\right) + t\left(2\overset{^}{i} + 2\overset{^}{j} - \overset{^}{k}\right), t \in R
Step-by-step solution
{L}_{1}: \overset{\to}{r} = \overset{^}{i} + \lambda\left(- \overset{^}{i} + 2\overset{^}{j} + 2\overset{^}{k}\right), \lambda \in R
Let a general point on line be
Also {L}_{2}: \overset{\to}{r} = \mu\left(2\overset{^}{i} - \overset{^}{j} + 2\overset{^}{k}\right)
let a general point on line be
Then \overset{\to}{AB} = \left(2\mu + \lambda - 1\right)\overset{^}{i} + \left(- \mu - 2\lambda\right)\overset{^}{j} + \left(2\mu - 2\lambda\right)\overset{^}{k}
Now, is perpendicular to both and
Hence, \overset{\to}{AB} ⋅ \left(- \overset{^}{i} + 2\overset{^}{j} + 2\overset{^}{k}\right) = 0 ( If and are perpendicular then )
& \overset{\to}{AB} ⋅ \left(2\overset{^}{i} - \overset{^}{j} + 2\overset{^}{k}\right) = 0
Using
\left(- 1\right)\left(2\mu + \lambda - 1\right) + \left(- \mu - 2\lambda\right)\left(2\right) + \left(2\mu - 2\lambda\right)2 = 0- 2\mu - \lambda + 1 - 2\mu - 4\mu + 4\mu - 4\lambda = 09\lambda = 1\lambda = \dfrac{1}{9} \ldots \left(i\right)( b )\overset{\to}{AB} ⋅ \left(2\overset{^}{i} - \overset{^}{j} + 2\overset{^}{k}\right) = 0⇒ \left(2\mu + \lambda - 1\right)\left(2\right) + \left(- \mu - 2\lambda\right)\left(- 1\right) + \left(2\mu - 2\lambda\right)\left(2\right) = 0⇒ 4\mu + 2\lambda - 2 + \mu + 2\lambda + 4\mu - 4\lambda = 0⇒ 9\mu = 2⇒ \mu = \dfrac{2}{9} \ldots \left(ii\right)\lambda\mu ⇒ A\left(\dfrac{8}{9}, \dfrac{2}{9}, \dfrac{2}{9}\right)B\left(\dfrac{4}{9}, - \dfrac{2}{9}, \dfrac{4}{9}\right)\overset{\to}{AB} = - \dfrac{2}{9}\left(2\overset{^}{i} + 2\overset{^}{j} - \overset{^}{k}\right)A\overset{\to}{AB}= \left(\dfrac{8}{9}\overset{^}{i} + \dfrac{2}{9}\overset{^}{j} + \dfrac{2}{9}\overset{^}{k}\right) + \gamma\left(- \dfrac{2}{9}\left(2\overset{^}{i} + 2\overset{^}{j} - \overset{^}{k}\right)\right)( B )\overset{\to}{r} = \dfrac{2}{9}\left(4\overset{^}{i} + \overset{^}{j} + \overset{^}{k}\right) + t\left(2\overset{^}{i} + 2\overset{^}{j} - \overset{^}{k}\right), t \in RB\overset{\to}{AB}( C )\overset{\to}{r} = \dfrac{2}{9}\left(2\overset{^}{i} - \overset{^}{j} + 2\overset{^}{k}\right) + t\left(2\overset{^}{i} + 2\overset{^}{j} - \overset{^}{k}\right), t \in RAB
Option :
\overset{\to}{r} = \dfrac{1}{3}\left(2\overset{^}{i} + \overset{^}{k}\right) + t\left(2\overset{^}{i} + 2\overset{^}{j} - \overset{^}{k}\right), t \in R