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) $\sqrt{\dfrac{2}{3}}$
(P)→(3) (Q)→(4) (R)→(1) (S)→(2)
(P)→(5) (Q)→(4) (R)→(1) (S)→(2)
(P)→(3) (Q)→(4) (R)→(1) (S)→(5)
(P)→(3) (Q)→(1) (R)→(4) (S)→(5)
Step-by-Step Solution
Key Concept: Equate parametric coordinates to find γ and intersection point; cross product of direction vectors gives normal
Parametrise: $L_1:(-11+t,-21+2t,-29+3t)$, $L_2:(-16+3s,-11+2s,-4+\gamma s)$.
Equating $x$: $t=3s-5$. Equating $y$: $t-s=5\Rightarrow s=5,t=10$.
Equating $z$: $3(10)-\gamma(5)=-29+29=25\Rightarrow30-5\gamma=25\Rightarrow\gamma=1$. → (P)→(3).
$R_1=L_1$ at $t=10$: $(-1,-1,1)=-\hat{i}-\hat{j}+\hat{k}$. → (R)→(1).
Directions: $\vec{d_1}=(1,2,3)$, $\vec{d_2}=(3,2,1)$.
$\vec{n}=\vec{d_1}\times\vec{d_2}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&2&3\\3&2&1\end{vmatrix}=(2-6,9-1,2-6)=(-4,8,-4)\propto(1,-2,1)$.
$\hat{n}=\dfrac{(1,-2,1)}{\sqrt{6}}=\dfrac{1}{\sqrt{6}}\hat{i}-\dfrac{2}{\sqrt{6}}\hat{j}+\dfrac{1}{\sqrt{6}}\hat{k}$. → (Q)→(4).
$\overrightarrow{OR_1}\cdot\hat{n}=(-1,-1,1)\cdot\dfrac{(1,-2,1)}{\sqrt{6}}=\dfrac{-1+2+1}{\sqrt{6}}=\dfrac{2}{\sqrt{6}}=\sqrt{\dfrac{2}{3}}$. → (S)→(5).
Answer: (P)→(3),(Q)→(4),(R)→(1),(S)→(5) → C.
Correct Answer: C