Question

Let l₁ and l₂ be the lines r₁ = λ(\(\hat{i}+\hat{j}+\hat{k}\)) and r₂ = (\(\hat{j}-\hat{k}\)) + μ (\(\hat{i}+\hat{k}\)), respectively. Let X be the set of all the planes H containing line l₁. For a plane H, let d (H) denote the smallest possible distance between the points of l₂ and H. Let H₀ be a plane in X for which d (H₀) is the maximum value of d (H ) as H varies over all planes in X . Match each entry in List-I to the correct entries in List-II.

List-I		List-II
(P)	The value of d (H0) is	(1)	\(\sqrt3\)
(Q)	The distance of the point (0,1,2) from H0 is	(2)	\(\frac{1}{\sqrt3}\)
(R)	The distance of origin from H0 is	(3)	0
(S)	The distance of origin from the point of intersection of planes y = z, x = 1, and H0 is	(4)	\(\sqrt2\)
		(5)	\(\frac{1}{\sqrt2}\)

 The correct option is: 

• A. (P)\(\rightarrow\)(2) (Q)\(\rightarrow\)(4) (R)\(\rightarrow\)(5) (S)\(\rightarrow\)(1)
• B. (P)\(\rightarrow\)(5) (Q)\(\rightarrow\)(4) (R)\(\rightarrow\)(3) (S)\(\rightarrow\)(1)
• C. (P)\(\rightarrow \)(2) (Q)\(\rightarrow\)(1) (R)\(\rightarrow\)(3) (S)\(\rightarrow\)(2)
• D. (P)\(\rightarrow\)(5) (Q)\(\rightarrow\)(1) (R)\(\rightarrow\)(4) (S)\(\rightarrow\)(2)

Answer 1

The Correct Option is B

The correct option is (B)
The normal vector of plane parallel l₁ and l₂ is
\(\begin{vmatrix} i &j  &k \\   1&1  &1 \\   1&0  &1  \end{vmatrix}=\hat{j}(1)-\hat{j}(1-1)+\hat{k}(-1)\)
\(=\hat{i}-\hat{j}\)
\(\therefore H_0:x-z=c|_{0,0,0}\)
\(\Rightarrow\) C = 0
\(H_0: x-z = 0\)
(P) d(H₀)=1 distance of point (0,1,-1) from H.
\(d=|\frac{0-(-1)}{\sqrt2}|=\frac{1}{\sqrt2}\therefore P\rightarrow 5\)
(Q) \(d=|\frac{0-2}{\sqrt2}|=\sqrt2\therefore Q\rightarrow 4\)
(R) \(d=|\frac{0}{\sqrt2}|=0\therefore S\rightarrow 3\)
(S)The point of intersection will be (1,1,1)
\(\therefore S\rightarrow 1\)
d=\(\sqrt{1+1+1}=\sqrt3\)