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JEE Advanced 2023Paper 1
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 ℓ1and{ℓ}_{1} and{ℓ}_{2}bethelinesbe the lines\overset{\to}{r}1 = \lambda\left(\overset{^}{i} + \overset{^}{j} + \overset{^}{k}\right)andand\overset{\to}{r}2 = \left(\overset{^}{j} - \overset{^}{k}\right) + \mu\left(\overset{^}{i} + \overset{^}{k}\right),respectively,Let, respectively, LetXbethesetofalltheplanesbe the set of all the planesHthatcontainthelinethat contain the line{ℓ}_{1}.Foraplane. For a planeH,let, letd\left(H\right)denotethesmallestpossibledistancebetweenthepointsofdenote the smallest possible distance between the points of{ℓ}_{2}andandH.Let. Let{H}_{0}beaplaneinbe a plane inXforwhichfor whichd\left({H}_{0}\right)isthemaximumvalueofis the maximum value ofd\left(H\right)asasHvariesoverallplanesinvaries over all planes inX.</p><p>MatcheachentryinList−ItothecorrectentriesinList−II.</p><tableborder="1"cellpadding="1"cellspacing="1"style="width:500px;"><tbody><tr><td></td><td>List−I</td><td></td><td>List−II</td></tr><tr><td>.</p><p>Match each entry in List-I to the correct entries in List-II.</p><table border="1" cellpadding="1" cellspacing="1" style="width: 500px;"> <tbody> <tr> <td> </td> <td>List-I</td> <td> </td> <td>List-II</td> </tr> <tr> <td>\left(P\right)</td><td>Thevalueof</td> <td>The value ofd\left({H}_{0}\right)is</td><td>(1)is</td> <td>\left(1\right) 3\sqrt{3} \left(Q\right)</td><td>Thedistanceofthepoint(0,1,2)</td> <td>The distance of the point \left(0, 1, 2\right) from H0{H}_{0} is \left(2\right)</td><td></td> <td>\dfrac{1}{\sqrt{3}}</td></tr><tr><td>(R)</td> </tr> <tr> <td>\left(R\right) The distance of origin from H0{H}_{0} is \left(3\right)</td><td></td> <td>0</td></tr><tr><td>(S)</td> </tr> <tr> <td>\left(S\right) The distance of origin from the point of
intersection of planes y=z,x=1y = z, x = 1 and H0{H}_{0} is \left(4\right)</td><td></td> <td>\sqrt{2}</td></tr><tr><td></td><td></td><td>(5)</td> </tr> <tr> <td> </td> <td> </td> <td>\left(5\right) 12\dfrac{1}{\sqrt{2}}

The correct option is

Options

  1. A(P)→(2)(Q)→(4)(R)→(5)(S)→(1)\left(P\right) \to \left(2\right)\left(Q\right) \to \left(4\right)\left(R\right) \to \left(5\right)\left(S\right) \to \left(1\right)
  2. B(P)→(5)(Q)→(4)(R)→(3)(S)→(1)\left(P\right) \to \left(5\right)\left(Q\right) \to \left(4\right)\left(R\right) \to \left(3\right)\left(S\right) \to \left(1\right)
  3. C(P)→(2)(Q)→(1)(R)→(3)(S)→(2)\left(P\right) \to \left(2\right)\left(Q\right) \to \left(1\right)\left(R\right) \to \left(3\right)\left(S\right) \to \left(2\right)
  4. D(P)→(5)(Q)→(1)(R)→(4)(S)→(2)\left(P\right) \to \left(5\right)\left(Q\right) \to \left(1\right)\left(R\right) \to \left(4\right)\left(S\right) \to \left(2\right)

Correct answer

B. (P)→(5)(Q)→(4)(R)→(3)(S)→(1)\left(P\right) \to \left(5\right)\left(Q\right) \to \left(4\right)\left(R\right) \to \left(3\right)\left(S\right) \to \left(1\right)

Step-by-step solution

Given,

H0{H}_{0} will be the plane containing the line ℓ1{ℓ}_{1} and parallel to ℓ2{ℓ}_{2}.

So, the normal vector of plane parallel to ℓ1andℓ2{ℓ}_{1} and {ℓ}_{2} 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}</p><p>Hence,theequationofplane</p><p>Hence, the equation of plane{H}_{0}willbe,</p><p>will be,</p><p>{H}_{0}: x - z = Cwhichpassesthroughorigin,</p><p>So,which passes through origin,</p><p>So,C = 0</p><p></p><p>∴ {H}_{0}: x - z = 0</p><p>Nowsolving,</p><p>(P)d(H0)=1</p><p>Now solving,</p><p>\left(P\right) d\left({H}_{0}\right) = 1 distance of point \left(0, 1, - 1\right)fromfromH.</p><p>.</p><p>d = \left|\dfrac{0 - \left(- 1\right)}{\sqrt{2}}\right| = \dfrac{1}{\sqrt{2}} ∴ P \to 5</p><p>(Q)d=∣0−22∣=2∴Q→4</p><p>\left(Q\right) d = \left|\dfrac{0 - 2}{\sqrt{2}}\right| = \sqrt{2} ∴ Q \to 4

\left(R\right) d = \left|\dfrac{0}{\sqrt{2}}\right| = 0 ∴ R \to 3</p><p>(S)</p><p>\left(S\right) Point of intersection will be of given planesy=z,x=1 and x−z=0y = z, x = 1 \text{ and } x - z = 0 will be, \left(1, 1, 1\right)</p><p>Hence,distance</p><p>Hence, distanced = \sqrt{1 + 1 + 1} = \sqrt{3} ∴ S \to 1</p><p></p><p>∴ Option (B) is correct.

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