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 S = { X ∈ R 3 : ( d

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 S = { X ∈ R 3 : ( dist ( X , P ) ) 2 − ( dist ( X , Q ) ) 2 = 50 } and T = { Y ∈ R 3 : ( dist ( Y , Q ) ) 2 − ( dist ( Y , P ) ) 2 = 50 } . ​ Then which of the following statements is (are) TRUE?

TestHub · Accepted Answer

​ S = { X : ( XP ) 2 − ( XQ ) 2 = 50 } T = { Y : ( YQ ) 2 − ( YP ) 2 = 50 } ​ for finding S ≡ X ( x , y , z ) and for T ≡ Y ( x , y , z ) ​ ( ( 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 } ​ 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 = ​ 10 105 − 5 ​ ​ = 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.

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

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