Let γ ∈ R be such that the lines L1 : x+11/1 = y+21/2 = z+29/3 and

Question

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 \(\vec {n}\)is	(2)	\(\sqrt{\frac{3}{2}}\)
(R)	\(\overrightarrow {OR}_1\) equals	(3)	1
(S)	A possible value of \(\overrightarrow {OR} _1.\) \(\hat {n}\) 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 

(A) (P) → (3) (Q) → (4) (R) → (1) (S) → (2) 

(B) (P) → (5) (Q) → (4) (R) → (1) (S) → (2) 

(C) (P) → (3) (Q) → (4) (R) → (1) (S) → (5) 

(D) (P) → (3) (Q) → (1) (R) → (4) (S) → (5)

Answer 1

Correct option is (C) (P) → (3) (Q) → (4) (R) → (1) (S) → (5) 

Vector parallel to the line \(L_1\left(\right. \text{say } \left.\vec{b}_1\right)=-\hat{i}+2 \hat{i}+3 \hat{k}\)

Normal vector of plane \((\vec {n})\) containing \(L_1\) and \(L_2\) will be perpendicular to both \(\vec {b _1}\) and \(\overrightarrow {AB}\)

\(\Rightarrow \vec{n}=p(\overrightarrow {A B} \times \vec{n})=p(5 \hat{i}-10 \hat{j}-25 \hat{k}) \times(i+2 \hat{j}+3 \hat{k})\)

\(=p(20 \hat{i}-40 \hat{j}+20 \hat{k})\)

\( \Rightarrow \hat{n}=\frac{1}{\sqrt{6}} \hat{i}-\frac{2}{\sqrt{6}} \hat{j}+\frac{1}{\sqrt{6}} \hat{k}\)

Now, vector parallel to \(L_2 (\text{say }\, \vec{b}_2 )\) is perpendicular to \(\vec{n} \Rightarrow \vec{b}_2 \cdot \vec{n}=0\)

\((3 \hat{i}+2 \hat{j}+\gamma \hat{k}) \cdot p(20 \hat{i}-40 \hat{j}+20 \hat{k})=0\)

\(\Rightarrow \gamma=1\)

Now, for point of intersection (POI)

\(L_1: \frac{x+11}{1}=\frac{y+21}{2}=\frac{z+29}{3}=\lambda \) and \(L_2: \frac{x+16}{3}=\frac{y+11}{2}=\frac{z+4}{\gamma}=u\)

Comparing x and y coordinates, \(-11 + \lambda = -16 + 3u \) and \(-21+2 \lambda=-11+2 u\)

\(\Rightarrow \lambda=10, u=5\)

\(\Rightarrow \text{POI i.e.}, \overrightarrow {OR}_1:(-\hat{i}-\hat{j}+\hat{k})\) and \(\overrightarrow {OR} \cdot \hat{n}=\sqrt{\frac{2}{3}}\)