JEE Advanced 2024 Paper 1 · Q16 · 3D Geometry

Question

Let $\gamma \in \mathbb{R}$ be such that the lines $L_1 : \dfrac{x + 11}{1} = \dfrac{y + 21}{2} = \dfrac{z + 29}{3}$ and $L_2 : \dfrac{x + 16}{3} = \dfrac{y + 11}{2} = \dfrac{z + 4}{\gamma}$ intersect. Let $R_1$ be the point of intersection of $L_1$ and $L_2$. Let $O = (0, 0, 0)$, and $\hat{n}$ denote a unit normal vector to the plane containing both the lines $L_1$ and $L_2$.

Match each entry in List-I to the correct entry in List-II.

List-I:
(P) $\gamma$ equals
(Q) A possible choice for $\hat{n}$ is
(R) $\overrightarrow{OR_1}$ equals
(S) A possible value of $\overrightarrow{OR_1} \cdot \hat{n}$ is

List-II:
(1) $-\hat{i} - \hat{j} + \hat{k}$
(2) $\sqrt{\dfrac{3}{2}}$
(3) $1$
(4) $\dfrac{1}{\sqrt{6}} \hat{i} - \dfrac{2}{\sqrt{6}} \hat{j} + \dfrac{1}{\sqrt{6}} \hat{k}$
(5) $\sqrt{\dfrac{2}{3}}$

SolveKar AI · Accepted Answer

Parametrize $L_1$: $(x,y,z)=(-11+t,-21+2t,-29+3t)$. Plug into $L_2$: $\dfrac{-11+t+16}{3}=\dfrac{-21+2t+11}{2}=\dfrac{-29+3t+4}{\gamma}$. First two: $\dfrac{t+5}{3}=\dfrac{2t-10}{2}\Rightarrow 2(t+5)=3(2t-10)\Rightarrow 2t+10=6t-30\Rightarrow t=10$. So $R_1=(-1,-1,1)$, and $\overrightarrow{OR_1}=-\hat{i}-\hat{j}+\hat{k}$. (R)$	o$(1). From third: $\dfrac{3t-25}{\gamma}=\dfrac{t+5}{3}=5$ at $t=10$, so $\dfrac{5}{\gamma}=5\Rightarrow \gamma=1$. (P)$	o$(3). Direction vectors: $\vec{d_1}=(1,2,3)$, $\vec{d_2}=(3,2,1)$. Normal $\vec{n}=\vec{d_1}	imes\vec{d_2}=(2-6,9-1,2-6)=(-4,8,-4)$. Unit: $\frac{1}{\sqrt{96}}(-4,8,-4)=\frac{1}{2\sqrt{6}}(-4,8,-4)=(-\frac{2}{\sqrt{6}},\frac{4}{\sqrt{6}},-\frac{2}{\sqrt{6}})$. The candidate $\frac{1}{\sqrt{6}}\hat{i}-\frac{2}{\sqrt{6}}\hat{j}+\frac{1}{\sqrt{6}}\hat{k}$ is $-\frac{1}{2}$ times this — same line. (Q)$	o$(4). $\overrightarrow{OR_1}\cdot\hat{n}=(-1)\cdot\frac{1}{\sqrt{6}}+(-1)\cdot\frac{-2}{\sqrt{6}}+1\cdot\frac{1}{\sqrt{6}}=\frac{2}{\sqrt{6}}=\sqrt{\frac{2}{3}}$. (S)$	o$(5). Option (C).

JEE Advanced 2024 Paper 1 · Q16 · 3D Geometry

Solution