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JEE Advanced 2019 · Introduction to Three Dimensional Geometry

Let L1L_1 and L2L_2 denote the lines r⃗=i^+λ(−i^+2j^+2k^)\vec{r} = \hat{i} + \lambda(-\hat{i} + 2\hat{j} + 2\hat{k}), λ∈R\lambda \in R and r⃗=μ(2i^−j^+2k^)\vec{r} = \mu(2\hat{i} - \hat{j} + 2\hat{k}), μ∈R\mu \in R respectively. If L3L_3 is a line which is perpendicular to both L1L_1 and L2L_2 and cuts both of them, then which of the following options describe(s) L3L_3?

Answers: A. r⃗=29(4i^+j^+k^)+t(2i^+2j^−k^)\vec{r} = \frac{2}{9}(4\hat{i} + \hat{j} + \hat{k}) + t(2\hat{i} + 2\hat{j} - \hat{k}), t∈Rt \in R · B. r⃗=29(2i^−j^+2k^)+t(2i^+2j^−k^)\vec{r} = \frac{2}{9}(2\hat{i} - \hat{j} + 2\hat{k}) + t(2\hat{i} + 2\hat{j} - \hat{k}), t∈Rt \in R · D. r⃗=13(2i^+k^)+t(2i^+2j^−k^)\vec{r} = \frac{1}{3}(2\hat{i} + \hat{k}) + t(2\hat{i} + 2\hat{j} - \hat{k}), t∈Rt \in R

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Let L₁ and L₂ denote the lines r⃗ = î + λ(-î + 2ĵ + 2k̂), λ ∈ R and r⃗ = μ(2î - ĵ + 2k̂), μ ∈ R respectively. If L₃ is a line which is perpendicular to both L₁ and L₂ and cuts both of them, then which of the following options describe(s) L₃? L1:r⃗=i^+λ(−i^+2j^+2k^),b⃗1=−i^+2j^+2k^L_1: \vec{r} = \hat{i} + \lambda(-\hat{i}+2\hat{j}+2\hat{k}), \quad \vec{b}_1 = -\hat{i}+2\hat{j}+2\hat{k} L2:r⃗=μ(2i^−j^+2k^),b⃗2=2i^−j^+2k^L_2: \vec{r} = \mu(2\hat{i}-\hat{j}+2\hat{k}), \quad \vec{b}_2 = 2\hat{i}-\hat{j}+2\hat{k} L3⊥L1, L3⊥L2, and L3 cuts bothL_3 \perp L_1, \ L_3 \perp L_2, \text{ and } L_3 \text{ cuts both} To find: which option(s) describe L3\text{To find: which option(s) describe } L_3 1. b⃗3=b⃗1×b⃗21.\ \vec{b}_3 = \vec{b}_1 \times \vec{b}_2 2. P=A+sb⃗1, Q=B+μb⃗2, with PQ∥b⃗32.\ P = A + s\vec{b}_1,\ Q = B + \mu\vec{b}_2, \text{ with } PQ \parallel \vec{b}_3 3. check each option’s point lies on L33.\ \text{check each option's point lies on } L_3 b⃗3=b⃗1×b⃗2=∣i^j^k^−1222−12∣\vec{b}_3 = \vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 2 & 2 \\ 2 & -1 & 2 \end{vmatrix} =6i^+6j^−3k^=3(2i^+2j^−k^)= 6\hat{i} + 6\hat{j} - 3\hat{k} = 3(2\hat{i}+2\hat{j}-\hat{k}) P=A+sb⃗1=(1−s, 2s, 2s),Q=B+μb⃗2=(2μ, −μ, 2μ)P = A + s\vec{b}_1 = (1-s,\ 2s,\ 2s), \quad Q = B + \mu\vec{b}_2 = (2\mu,\ -\mu,\ 2\mu) PQ⃗=Q−P=k(2, 2, −1)\vec{PQ} = Q - P = k(2,\ 2,\ -1) 2μ−1+s=2k,−μ−2s=2k,2μ−2s=−k2\mu - 1 + s = 2k, \quad -\mu - 2s = 2k, \quad 2\mu - 2s = -k μ+s=13, μ=2s⇒s=19, μ=29\mu + s = \tfrac{1}{3}, \ \mu = 2s \Rightarrow s = \tfrac{1}{9}, \ \mu = \tfrac{2}{9} P(89,29,29),Q(49,−29,49)P\left(\tfrac{8}{9},\tfrac{2}{9},\tfrac{2}{9}\right), \quad Q\left(\tfrac{4}{9},-\tfrac{2}{9},\tfrac{4}{9}\right) (1): 29(4,1,1)=(89,29,29)=P ✓(1): \ \tfrac{2}{9}(4,1,1) = \left(\tfrac{8}{9},\tfrac{2}{9},\tfrac{2}{9}\right) = P \ \text{✓} (2): 29(2,−1,2)=(49,−29,49)=Q ✓(2): \ \tfrac{2}{9}(2,-1,2) = \left(\tfrac{4}{9},-\tfrac{2}{9},\tfrac{4}{9}\right) = Q \ \text{✓} (3): r=t(2,2,−1) passes through O(0,0,0)(3):\ r = t(2,2,-1) \text{ passes through } O(0,0,0) At x=0 on L3: y=−23≠0 ✗\text{At } x=0 \text{ on } L_3: \ y = -\tfrac{2}{3} \neq 0 \ \text{✗} (4): 13(2,0,1)=(23,0,13)(4): \ \tfrac{1}{3}(2,0,1) = \left(\tfrac{2}{3},0,\tfrac{1}{3}\right) =P−19(2,2,−1)∈L3 ✓= P - \tfrac{1}{9}(2,2,-1) \in L_3 \ \text{✓} Correct options: (1), (2), (4)\boxed{\text{Correct options: } (1),\ (2),\ (4)}

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