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
(Q)	The distance of the point (0,1,2) from H0 is
(R)	The distance of origin from H0 is
(S)	The distance of origin from the point of intersection of planes y = z, x = 1, and H0 is
	(5)

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

Answer: (B)

(P)$\rightarrow$(5) (Q)$\rightarrow$(4) (R)$\rightarrow$(3) (S)$\rightarrow$(1)

Answer 2

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=$\sqrt{1+1+1}=\sqrt3$