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JEE Advanced Mathematics Three Dimensional Geometry 2024 JEE Advanced 2024 (Paper 1)

JEE Advanced Mathematics Question (2024) — Solution

Question

Let R ^3 denote the three-dimensional space. Take two points P=(1,2,3) and Q=(4,2,7). Let dist (X, Y) denote the distance between two points X and Y in R ^3. Let gathered S= \ X R ^3:( dist (X, P))^2-( dist (X, Q))^2=50 \ and \ = \ Y R ^3:( dist (Y, Q))^2-( dist (Y, P))^2=50 \ . gathered Then which of the following statements is (are) TRUE?

Options

  1. A. There is a triangle whose area is 1 and all of whose vertices are from S.
  2. B. There are two distinct points L and M in T such that each point on the line segment L M is also in T.
  3. C. There are infinitely many rectangles of perimeter 48 , two of whose vertices are from S and the other two vertices are from T.
  4. D. There is a square of perimeter 48 , two of whose vertices are from S and the other two vertices are from T.

Answer

D. There is a square of perimeter 48 , two of whose vertices are from S and the other two vertices are from T.

Step-by-step solution

aligned & S = \ X :( XP )^2-( XQ )^2=50 \ \\& T = \ Y :( YQ )^2-( YP )^2=50 \ aligned for finding S X(x, y, z) and for T Y(x, y, z) aligned & ((x-1)^2+(y-1)^2+( z -1)^2 )- (( x -4)^2+( y -2)^2+( z -7)^2 )=50 \\& S =\ ( x , y , z ): 6 x +8 z =105\ \\& T =\ ( x , y , z ): 6 x +8 z =5\ aligned Since S and T both are plane ; (1) There exist a triangle in plane S whose area =1 (always) (2) L ~\&~ M lies on plane T , hence line segment joining L \& M will lie on plane T . (3) Distance between S \& ~T d = | 105-5 10 |=10 Hence for rectangle of perimeter 48 can exist. (4) For Square There will be infinite such rectangle possible. Hence Answers 1,2,3,4 are correct.

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