Match the following. List I List II (A) Let P be the plane passing through the point ( 2 , 1 , − 1 ) and perpendicular to the line of intersection of the planes 2 x + y − z = 3 and x + 2 y + z = 2 . T

Question

Match the following.

List I	List II
(A) Let $P$ be the plane passing through the point $(2, 1, - 1)$ and perpendicular to the line of intersection of the planes $2 x + y - z = 3$ and $x + 2 y + z = 2$ . Then the distance from the point $(3, 2, 2)$ to the plane $P$ is	(P) $\frac{5}{3}$
(B) If $lim_{n \to \infty} \sum_{i = 1}^{n} tan^{- 1} (\frac{1}{2 i ^{2}}) = t$ , then $tan t$ is	(Q) $1$
(C) The sides $a, b, c$ of a triangle $A BC$ are in A.P. If $cos θ_{1} = \frac{a}{b + c}$ , $cos θ_{2} = \frac{b}{a + c}$ , $cos θ_{3} = \frac{c}{a + b}$ , then $tan^{2} \frac{θ _{1}}{2} + tan^{2} \frac{θ _{3}}{2}$ is	(R) $\frac{2}{3}$
(D) Let $L$ be the line passing through the point $(0, 1, 0)$ and perpendicular to the plane $x + 2 y + 2 z = 0$ . Then the distance from the point $(0, 0, 0)$ to the line $L$ is	(S) $0$

TestHub · Accepted Answer

(A) The direction of the line of intersection is n 1 ​ × n 2 ​ = ( 2 , 1 , − 1 ) × ( 1 , 2 , 1 ) = ( 3 , − 3 , 3 ) , so the plane through ( 2 , 1 , − 1 ) is x − y + z = 0 . Distance of ( 3 ​ , 2 , 2 ) is 3 ​ ∣ 3 ​ − 2 + 2∣ ​ = 1 , so (A) → (Q). (B) tan − 1 2 i 2 1 ​ = tan − 1 2 i − 1 1 ​ - tan − 1 2 i + 1 1 ​ , so the sum telescopes to tan − 1 1 = 4 π ​ and tan t = 1 , so (B) → (Q). (C) tan 2 2 θ 1 ​ ​ = 1 + c o s θ 1 ​ 1 − c o s θ 1 ​ ​ = a + b + c b + c − a ​ and tan 2 2 θ 3 ​ ​ = a + b + c a + b − c ​ . The sum is a + b + c 2 b ​ = 3 b 2 b ​ = 3 2 ​ , so (C)→(R). (D) The direction of L is (1,2,2). With OP = ( 0 , 1 , 0 ) , OP × ( 1 , 2 , 2 ) = ( 2 , 0 , − 1 ) , so the distance is 3 5 ​ ​ , so (D) → (P). Hence A-Q, B-Q, C-R, D-P.

Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

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