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Example 6 - Find equation of set of points P PA2 + PB2 = 2k2

Example 6 - Chapter 12 Class 11 Introduction to Three Dimensional Geometry - Part 2
Example 6 - Chapter 12 Class 11 Introduction to Three Dimensional Geometry - Part 3 Example 6 - Chapter 12 Class 11 Introduction to Three Dimensional Geometry - Part 4

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Example 6 Find the equation of set of points P such that PA2 + PB2 = 2k2, where A and B are the points (3, 4, 5) and (–1, 3, –7), respectively. Given A (3, 4, 5) & B ( – 1, 3, −7) Let the co-ordinates of point P be (x, y, z) We need to find equation of set of point P (x, y, z) Such that PA2 + PB2 = 2k2 First, we calculate (PA)2, (PB)2 Calculating (PA)2 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))2 (PA)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 – (5) (z) = 9 + x2 – 6x + 16 + y2 – 8y + 25 + z2 – 10z = x2 + y2 + z2 – 6x – 8y –10z + 9 + 16 + 25 = x2 + y2 + z2 – 6x – 8y –10z + 50 Calculating (PB)2 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) Squaring both sides (PB)2 = (√((−1)^2 (1+x)^2+(3−y)2+(−1)(7+z)2) " " )^2 = (1)2 + (x)2 + 2(1)(x) + (3)2 + (y)2 – 2(3)(y) + (7)2 + (z)2 + 2(7) (z) = 1 + x2 + 2x + 9 + y2 – 6y + 49 + z2 + 14z Putting value of (PA)2 & (PB)2 in (1) (PA)2 + (PB)2 = 2k2 (x2 + y2 + z2 – 6x – 8y – 10z + 50) + (x2 + y2 + z2 + 2x – 6y + 14z + 59) = 2k2 2x2 + 2y2 + 2z2 – 4x – 14y + 4z + 50 + 59 = 2k2 2x2 + 2y2 + 2z2 – 4x – 14y + 4z = 2k2 – 50 – 59 2x2 + 2y2 + 2z2 – 4x – 14y + 4z = 2k2 – 109 which is the required equation

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.