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JEE Advanced 2023 · Introduction to Three Dimensional Geometry

Let ℓ1\ell_{1} and ℓ2\ell_{2} be the lines r⃗1=λ(i^+j^+k^)\vec{r}_{1} = \lambda(\hat{i}+\hat{j}+\hat{k}) and r⃗2=(j^−k^)+μ(i^+k^)\vec{r}_{2} = (\hat{j}-\hat{k}) + \mu(\hat{i}+\hat{k}), respectively. Let X be the set of all the planes H that contain the line ℓ1\ell_{1}. For a plane H, let d(H)d(H) denote the smallest possible distance between the points of ℓ2\ell_{2} and H. Let H0H_{0} be a plane in X for which d(H0)d(H_{0}) is the maximum value of d(H)d(H) as H varies over all planes in X. Match each entry in List-I to the correct entries in List-II. | List-I | List-II | | --- | --- | | (P) The value of d(H0)d(H_{0}) is | (1) 3\sqrt{3} | | (Q) The distance of the point (0,1,2)(0, 1, 2) from H0H_{0} is | (2) 13\frac{1}{\sqrt{3}} | | (R) The distance of origin from H0H_{0} is | (3) 00 | | (S) The distance of origin from the point of intersection of planes y=zy = z, x=1x = 1 and H0H_{0} is | (4) 2\sqrt{2} | | | (5) 12\frac{1}{\sqrt{2}} | The correct option is:

Answer: B. (P)→(5), (Q)→(4), (R)→(3), (S)→(1)\text{(P)} \to \text{(5)},\ \text{(Q)} \to \text{(4)},\ \text{(R)} \to \text{(3)},\ \text{(S)} \to \text{(1)}

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Let ell₁ and ell₂ be the lines r⃗₁ = λ(î+ĵ+k̂) and r⃗₂ = (ĵ-k̂) + μ(î+k̂), respectively. Let X be the set of all the planes H that contain the line ell₁. For a plane H, let d(H) denote the smallest possible distance between the points of ell₂ and H. Let H₀ be a plane in X for which d(H₀) is the maximum value of d(H) as H varies over all planes in X. Match each entry in List-I to the correct entries in List-II. | List-I | List-II | | -- ℓ1: r⃗=λ(i^+j^+k^)⇒through (0,0,0), d⃗1=(1,1,1)\ell_1:\ \vec{r}=\lambda(\hat{i}+\hat{j}+\hat{k}) \Rightarrow \text{through } (0,0,0), \ \vec{d}_1=(1,1,1) ℓ2: r⃗=(j^−k^)+μ(i^+k^)⇒through (0,1,−1), d⃗2=(1,0,1)\ell_2:\ \vec{r}=(\hat{j}-\hat{k})+\mu(\hat{i}+\hat{k}) \Rightarrow \text{through } (0,1,-1), \ \vec{d}_2=(1,0,1) X={H:ℓ1⊂H},d(H)=min⁡dist(ℓ2,H)X=\{H:\ell_1\subset H\}, \quad d(H)=\min \text{dist}(\ell_2, H) H0∈X maximises d(H)H_0\in X \text{ maximises } d(H) Match (P),(Q),(R),(S) with (1)–(5)\text{Match (P),(Q),(R),(S) with (1)--(5)} 1. H: n⃗⋅r⃗=0,n⃗⊥d⃗11.\ H:\ \vec{n}\cdot\vec{r}=0,\quad \vec{n}\perp\vec{d}_1 2. d(H)=0 unless ℓ2∥H2.\ d(H)=0 \text{ unless } \ell_2\parallel H 3. maximise d(H)⇒H0, then (P),(Q),(R),(S)3.\ \text{maximise } d(H)\Rightarrow H_0, \text{ then (P),(Q),(R),(S)} H: n⃗⋅r⃗=0 (through O)H:\ \vec{n}\cdot\vec{r}=0 \ \text{(through } O) n⃗⊥d⃗1⇒n1+n2+n3=0\vec{n}\perp\vec{d}_1 \Rightarrow n_1+n_2+n_3=0 d⃗2⋅n⃗=n1+n3\vec{d}_2\cdot\vec{n}=n_1+n_3 d⃗2⋅n⃗≠0⇒ℓ2 meets H⇒d(H)=0\vec{d}_2\cdot\vec{n}\ne 0 \Rightarrow \ell_2 \text{ meets } H \Rightarrow d(H)=0 d(H)≠0⇒n1+n3=0d(H)\ne 0 \Rightarrow n_1+n_3=0 n2=0,n⃗∥(1,0,−1)n_2=0,\quad \vec{n}\parallel(1,0,-1) H0: x−z=0H_0:\ x-z=0 d(H0)=∣n⃗⋅OA→∣∣n⃗∣,A(0,1,−1), n⃗=(1,0,−1)d(H_0)=\dfrac{|\vec{n}\cdot\overrightarrow{OA}|}{|\vec{n}|},\quad A(0,1,-1),\ \vec{n}=(1,0,-1) n⃗⋅OA→=1,∣n⃗∣=2\vec{n}\cdot\overrightarrow{OA}=1, \quad |\vec{n}|=\sqrt2 d(H0)=12⇒(P)→(5)d(H_0)=\dfrac{1}{\sqrt2} \Rightarrow \text{(P)}\to\text{(5)} dist((0,1,2),H0)=∣0−2∣2\text{dist}\big((0,1,2),H_0\big)=\dfrac{|0-2|}{\sqrt2} =22=2⇒(Q)→(4)=\dfrac{2}{\sqrt2}=\sqrt2 \Rightarrow \text{(Q)}\to\text{(4)} O(0,0,0)∈H0⇒dist(O,H0)=0O(0,0,0)\in H_0 \Rightarrow \text{dist}(O,H_0)=0 ⇒(R)→(3)\Rightarrow \text{(R)}\to\text{(3)} y=z, x=1, x=z⇒(1,1,1)y=z,\ x=1,\ x=z \Rightarrow (1,1,1) dist from O=1+1+1=3⇒(S)→(1)\text{dist from } O=\sqrt{1+1+1}=\sqrt3 \Rightarrow \text{(S)}\to\text{(1)} (P)→(5), (Q)→(4), (R)→(3), (S)→(1)\text{(P)}\to(5),\ \text{(Q)}\to(4),\ \text{(R)}\to(3),\ \text{(S)}\to(1) Answer: Option (2)\boxed{\text{Answer: Option (2)}}

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