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JEE Advance - Mathematics (2024 - Paper 1 Online - No. 16)

Let $\gamma \in \mathbb{R}$ be such that the lines $L_1: \frac{x+11}{1}=\frac{y+21}{2}=\frac{z+29}{3}$ and $L_2: \frac{x+16}{3}=\frac{y+11}{2}=\frac{z+4}{\gamma}$ 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 List-II
(P) $\gamma$ equals (1) $-\hat{i} - \hat{j} + \hat{k}$
(Q) A possible choice for $\hat{n}$ is (2) $\sqrt{\frac{3}{2}}$
(R) $\overrightarrow{OR_1}$ equals (3) $1$
(S) A possible value of $\overrightarrow{OR_1} \cdot \hat{n}$ is (4) $\frac{1}{\sqrt{6}} \hat{i} - \frac{2}{\sqrt{6}} \hat{j} + \frac{1}{\sqrt{6}} \hat{k}$
(5) $\sqrt{\frac{2}{3}}$

The correct option is :
$(\mathrm{P}) \rightarrow(3) \quad(\mathrm{Q}) \rightarrow(4) \quad(\mathrm{R}) \rightarrow(1) \quad(\mathrm{S}) \rightarrow(2)$
$(\mathrm{P}) \rightarrow(5) \quad(\mathrm{Q}) \rightarrow(4) \quad(\mathrm{R}) \rightarrow(1) \quad(\mathrm{S}) \rightarrow(2)$
$(\mathrm{P}) \rightarrow(3) \quad$ (Q) $\rightarrow(4) \quad(\mathrm{R}) \rightarrow(1) \quad$ (S) $\rightarrow(5)$
$(\mathrm{P}) \rightarrow(3) \quad(\mathrm{Q}) \rightarrow(1) \quad(\mathrm{R}) \rightarrow(4) \quad$ (S) $\rightarrow(5)$

Explanation

$$\begin{aligned} & L_1: \frac{x+11}{1}=\frac{y+21}{2}=\frac{z+29}{3}=a \\ & L_2: \frac{x+16}{3}=\frac{y+11}{2}=\frac{z+4}{\gamma}=b \end{aligned}$$

$$\begin{aligned} & x=a-11=3 b-16 \Rightarrow a-3 b=-5 \quad \text{.... (1)}\\ & y=2 a-21=2 b-11 \Rightarrow 2 a-2 b=10 \quad \text{.... (2)}\\ & z=3 a-29=b r-4 \Rightarrow 3 a-b y=25 \quad \text{.... (3)} \end{aligned}$$

$$\begin{aligned} & \text { from (1) & (2) } \\ & a=10, b=5 \end{aligned}$$

Now from (3)

$$\begin{aligned} & 3(10)-5 \gamma=25 \quad \therefore \gamma=1 \\ & \mathrm{R}_1 \equiv(-1,-1,1) \\ & \mathrm{OR}_1=-\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}} \end{aligned}$$

$$\begin{aligned} & \vec{n}=\left|\begin{array}{lll} i & j & k \\ 1 & 2 & 3 \\ 3 & 2 & 1 \end{array}\right|=-4 \hat{i}-(-8) \hat{j}-4 \hat{k} \\ & \vec{n}=-4 \hat{i}+8 \hat{j}+4 \hat{k}=-4(\hat{i}-2 \hat{j}+\hat{k}) \\ & \hat{n}= \pm \frac{4(\hat{i}-2 \hat{j}+\hat{k})}{4 \sqrt{6}}= \pm \frac{(\hat{i}-2 \hat{j}+\hat{k})}{\sqrt{6}} \\ & \overrightarrow{O R} \cdot \hat{n}= \pm(-\hat{i}-\hat{j}+\hat{k})\left(\frac{\hat{i}-2 \hat{j}+\hat{k}}{\sqrt{6}}\right)= \pm \frac{2}{\sqrt{6}}= \pm \sqrt{\frac{4}{6}}= \pm \sqrt{\frac{2}{3}} \end{aligned}$$

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