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Three Dimensional Geometry

Section: Mathematics

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Question : 31 of 37

Marks: +1, -0

Match the statements in Column I with those in Column II:

Column I	Column II
(A) A line from the origin meets the lines $\frac{x-2}{1}$ = $\frac{y-1}{-2}$ = $\frac{z+1}{1}$ and $\frac{x-\frac{8}{3}}{2}$ = $\frac{y+3}{-1}$ = $\frac{z-1}{1}$ at P and Q, respectively. If length PQ = d, then $d^2$ is	(P) - 4
(B) The values of x satisfying $\tan^{-1}(x+3)-\tan^{-1}(x-3)$ = $\sin^{-1}\left(\frac{3}{5}\right)$ are	(Q) 0
(C) Non-zero vectors $\overset{\rightarrow}{a},\overset{\rightarrow}{b}$ and $\overset{\rightarrow}{c}$ satisfy $\overset{\rightarrow}{a}\cdot\overset{\rightarrow}{b}$ = 0, $\left(\overset{\rightarrow}{b}-\overset{\rightarrow}{a}\right)\cdot\left(\overset{\rightarrow}{b}+\overset{\rightarrow}{c}\right)$ = 0 and $2\left\|\overset{\rightarrow}{b}+\overset{\rightarrow}{c}\right\|$ = $\left\|\overset{\rightarrow}{b}-\overset{\rightarrow}{a}\right\|$ . If $\overset{\rightarrow}{a}$ = $\mu\overset{\rightarrow}{b}+\overset{\rightarrow}{4c}$ , then the possible values of µ are	(R) 4
(D) Let f be the function on [–π, π] given by f (0) = 9 and f (x) = $\frac{\sin\left(\frac{9x}{2}\right)}{\sin\left(\frac{x}{2}\right)}$ for x ≠ 0. The value of $\frac{2}{\pi}\int\limits_{-\pi}^{\pi}f(x)\,dx$ is	(S) 5
	(T) 6

[JEE Adv 2010 P2]

(A)→(P), (B)→(P), (R); (C)→(R), (S); (D)→(R)
(A)→(R), (B)→(P), (R); (C)→(Q), (S); (D)→(P)
(A)→(T), (B)→(P), (R); (C)→(Q), (S); (D)→(R)
(A)→(S), (B)→(P), (R); (C)→(Q), (S); (D)→(T)

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