- Exam
- JEE Advanced
- Year
- 2023
- Paper
- Paper 1
- Subject
- Mathematics
- Chapter
- Three Dimensional Geometry
JEE Advanced 2023 · Paper 1 · Mathematics · Three Dimensional Geometry
Let ℓ _ 1 and ℓ _ 2 be the lines → r 1 = λ( ^ i + ^ j + ^ k ) and →...
Let {ℓ}_{2}\overset{\to}{r}1 = \lambda\left(\overset{^}{i} + \overset{^}{j} + \overset{^}{k}\right)\overset{\to}{r}2 = \left(\overset{^}{j} - \overset{^}{k}\right) + \mu\left(\overset{^}{i} + \overset{^}{k}\right)XH{ℓ}_{1}Hd\left(H\right){ℓ}_{2}H{H}_{0}Xd\left({H}_{0}\right)d\left(H\right)HX\left(P\right)d\left({H}_{0}\right)
intersection of planes and is
The correct option is
Options
- A
- B
- C
- D
Correct answer
B.
Step-by-step solution
Given,
will be the plane containing the line and parallel to .
So, the normal vector of plane parallel to is given by,
\left|\begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ 1 & 1 & 1 \\ 1 & 0 & 1\end{cases}\right| = \overset{^}{j}\left(1\right) - \overset{^}{j}\left(1 - 1\right) + \overset{^}{k}\left(- 1\right)= \overset{^}{i} - \overset{^}{k}{H}_{0}{H}_{0}: x - z = CC = 0∴ {H}_{0}: x - z = 0 distance of point \left(0, 1, - 1\right)Hd = \left|\dfrac{0 - \left(- 1\right)}{\sqrt{2}}\right| = \dfrac{1}{\sqrt{2}} ∴ P \to 5
\left(R\right) d = \left|\dfrac{0}{\sqrt{2}}\right| = 0 ∴ R \to 3 Point of intersection will be of given planes will be, \left(1, 1, 1\right)d = \sqrt{1 + 1 + 1} = \sqrt{3} ∴ S \to 1∴ Option (B) is correct.