Match the following. List-I List-II (A) The distance of the point ( 1 , − 2 , 3 ) from the plane x − y + z − 5 = 0 measured parallel to 2 x = 3 y = − 6 z − 1 is (p) 0 (B) If the straight lines 1

Question

Match the following.

List-I	List-II
(A) The distance of the point $(1, - 2, 3)$ from the plane $x - y + z - 5 = 0$ measured parallel to $\frac{x}{2} = \frac{y}{3} = \frac{z - 1}{- 6}$ is	(p) $0$
(B) If the straight lines $\frac{x - 2}{1} = \frac{y - 3}{1} = \frac{4 - z}{K}$ and $\frac{x - 1}{K} = \frac{y - 4}{2} = \frac{z - 5}{1}$ intersect, then $K$ is equal to	(q) $1$
(C) The shortest distance between any two opposite edges of the tetrahedron formed by the planes $y + z = 0$ , $z + x = 0$ , $x + y = 0$ and $x + y + z = 6$ is	(r) $- 3$
(D) If $θ$ is the angle between the line $x = y = z$ and the plane $x + y + z = 4$ , then $tan \frac{θ}{2}$ is	(s) $2$

TestHub · Accepted Answer

(A) Point on the parallel line: ( 1 + 2 λ , − 2 + 3 λ , 3 − 6 λ ) ; substituting in the plane: 1 − 7 λ = 0 , λ = 7 1 ​ ; distance = 7 1 ​ ⋅ 4 + 9 + 36 ​ = 1 . (B) Coplanarity: ​ 1 ​ − 1 ​ − 11 ​ 1 ​ − KK ​ 2 ​ 1 ​ ​ = K 2 + 3 K = 0 , so K = 0 or K = − 3 . (C) Vertices O ( 0 , 0 , 0 ) , A ( 6 ​ , 6 ​ , − 6 ​ ) , B ( 6 ​ , − 6 ​ , 6 ​ ) , C ( − 6 ​ , 6 ​ , 6 ​ ) . For opposite edges O A (direction ( 1 , 1 , − 1 ) ) and BC (direction ( − 1 , 1 , 0 ) ): common normal ( 1 , 1 , 2 ) , distance = 6 ​ ∣ ( 6 ​ , − 6 ​ , 6 ​ ) ⋅ ( 1 , 1 , 2 ) ∣ ​ = 6 ​ 2 6 ​ ​ = 2 ; the other opposite-edge pairs give the same. (D) The line's direction (1, 1, 1) is the plane's normal: θ = 2 π ​ , tan 2 θ ​ = tan 4 π ​ = 1 .

Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

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