Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
27.1k views
in Three-dimensional geometry by (54.8k points)

A variable plane cuts the x-axis, y-axis and z-axis at the points A, B and C, respectively, such that the volume of the tetrahedron OABC remains constant equal to 32 cubic unit and O is the origin of the coordinate system. 

Column I  Column II
(A) The locus of the centroid of the tetrahedron is  (p) xyz = 24 
(B) the locus of the point equidistant from O, A, B and C is  (q) (x2 + y2 + z2)3 = 192xyz
(C) The length of the foot of perpendicular from origin to the plane is  (r) xyz = 3 
(D) If PA, PB and PC are mutually perpendicular then the locus of P is (s) (x2 + y2 + z2) 3 = 1536xyz

1 Answer

+1 vote
by (52.5k points)
selected by
 
Best answer

Answer is (A) → (r), (B) → (p), (C) → (q), (D) → (s)

Given abc/6 = 32, where A, B, C are, respectively, (a, 0, 0), (0, b, 0), (0, 0, c).

(A) Centroid of tetrahedron (x, y z) ≡ (a/4, b/4, c/4)

⇒ a = 4x, b = 4y, c = 4z

Therefore,

64xyz = 32 × 6 ⇒ xyz = 3

(B) Equidistant point (x, y, z) ≡ (a/2, b/2, c/2)

⇒ a = 2x, b = 2y, c = 2z

Therefore,

8abg = 32 × 6 ⇒ xyz = 24

(C) The equation of the plane is x/a + y/b + z/c = 1

Therefore, foot of perpendicular from the origin ≡ (x, y, z)

Therefore, 

xyz = 6 × 32

⇒ (x2 + y2 + z2)3 = 192 × 8xyz = 1536xyz

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

...