Match the following. List-I List-II (A) The volume of the tetrahedron whose vertices are A ( 3 , 7 , 4 ) , B ( 5 , − 2 , 3 ) , C ( − 4 , 5 , 6 ) and D ( 1 , 2 , 3 ) . (P) 1 (B) The perpendicular dista

Question

Match the following.

List-I	List-II
(A) The volume of the tetrahedron whose vertices are $A (3, 7, 4)$ , $B (5, - 2, 3)$ , $C (- 4, 5, 6)$ and $D (1, 2, 3)$ .	(P) $1$
(B) The perpendicular distance between $2 x + 2 y - z + 1 = 0$ and $x + y - \frac{z}{2} + 2 = 0$ .	(Q) $0.74$
(C) A plane passes through $(1, - 2, 1)$ and is perpendicular to two planes $2 x - 2 y + z = 0$ and $x - y + 2 z = 4$ . The distance of the plane from the point $(1, 2, 2)$ is.	(R) $15.33$
(D) A line is perpendicular to $x + 2 y + 2 z = 0$ and passes through $(0, 1, 0)$ . The perpendicular distance of this line from $(0, 0, 0)$ is.	(S) $2.82$

TestHub · Accepted Answer

(A) Volume = 6 1 ​ ​ 3 ​ 7 ​ 4 ​ 15 ​ − 2 ​ 3 ​ 1 − 4 ​ 5 ​ 6 ​ 11 ​ 2 ​ 3 ​ 1 ​ ​ = 6 92 ​ = 15.33 → R . (B) The planes are parallel □ x + y − 2 z ​ + 2 = 0 ≡ 2 x + 2 y − z + 4 = 0 ): distance = 4 + 4 + 1 ​ ∣4 − 1∣ ​ = 1 → P. (C) Normal of the required plane is ( 2 , − 2 , 1 ) × ( 1 , − 1 , 2 ) = ( − 3 , − 3 , 0 ) ∼ ( 1 , 1 , 0 ) : plane x + y = λ through ( 1 , − 2 , 1 ) gives λ = − 1 , i.e. x + y + 1 = 0 ; distance from ( 1 , 2 , 2 ) = 2 ​ 4 ​ = 2 2 ​ = 2.82 → S . (D) The line through ( 0 , 1 , 0 ) along ( 1 , 2 , 2 ) : distance from origin = ∣ ( 1 , 2 , 2 ) ∣ ∣ ( 0 , 1 , 0 ) × ( 1 , 2 , 2 ) ∣ ​ = 3 ∣ ( 2 , 0 , − 1 ) ∣ ​ = 3 5 ​ ​ = 0.74 → Q .

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

Options:

Doubts & Discussion