# Misc 6 - Chapter 12 Class 11 Introduction to Three Dimensional Geometry

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Misc 6 If A and B be the points (3, 4, 5) and ( 1, 3, 7), respectively, find the equation of the set of points P such that PA2 + PB2 = k2, where k is a constant. Given points A (3, 4, 5) & B ( 1, 3, 7) Let Point P be (x, y, z,) We need to point equation of points P, such that PA2 + PB2 = k2 Calculating PA2 P (x, y, z) , A (3, 4, 5) PA = x2 x1 2+ y2 y1 2+ z2 z1 2 Here, x1 = x, y1 = y, z1 = z x2 = 3, y2 = 4, z2 = 5 PA = 3 x 2+ 4 y 2+ 5 z 2 Squaring both sides (PA)2 = 3 x 2+ 4 y 2+ 5 z 2 = (3 x)2 + (4 y)2 + (5 z)2 = (3)2 + (x)2 2(3) (x) + (4)2 + y2 2(4)(y) + (5)2 + (z)2 2(5)(z) = 9 + x2 6x + 16 + y2 8y + 25 + z2 10z = x2 + y2 6x 8y 10z + 9 + 16 + 25 = x2 + y2 + z2 6x 8y 10z + 50 Calculating PB2 P (x, y, z) , B ( 1, 3 , 7) PB = x2 x1 2+ y2 y1 2+ z2 z1 2 Here, x1 = x, y1 = y, z1 = z x2 = 1, y2 = 3, z2 = 7 PB = 1 x 2+ 3 y 2+ 7 z 2 PB = 1 2 1+ 2+ 3 y 2+ 1 2 7 z 2 PB = 1+ 2+ 3 y 2+ 7 z 2 Squaring both sides (PB)2 = 1+ 2+ 3 y 2+ 7+z 2 2 (PB)2 = (1 + x)2 + (3 y)2 + (7 + z)2 (PB)2 = (1)2 + x2 + 2(1)(x) + (3)2 + (y)2 2(3)(y) + (7)2 + z2 + 2(7)(z) = 1 + x2 + 2x + 9 + y2 6y + 49 + y2 + +14z = x2 + y2 + z2 + 2x 6y + 14z + 1 + 9 + 49 = x2 + y2 + z2 + 2x 6y + 14z + 59 Putting value of (PA)2 & (PB)2 in (1) (PA)2 + (PB)2 = k2 (x2 + y2 + z2 6x 8y 10z + 50) + (x2 + y2 + z2 + 2x 6y + 14z + 59) = k2 2x2 + 2y2 + 2z2 6x + 2x 8y 6y 10z + 4z + 50 + 59 = 52 2x2 + 2y2 + 2z2 4x 14y 6z + 109 = 52 2(x2 + y2 + z2) 4x 14y 6z + 109 52 = 0 Thus the required equation is 2(x2 + y2 + z2) 4x 14y 6z + 109 52 = 0

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.