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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+𝑥)2+(3−y)2+(−1)2(7−z)2) PB = √((1+𝑥)2+(3−y)2+(7−z)2) Squaring both sides (PB)2 = (√((1+𝑥)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 + z2 + 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 + 14z + 50 + 59 = k2 2x2 + 2y2 + 2z2 – 4x – 14y + 4z + 109 = k2 2(x2 + y2 + z2) – 4x – 14y + 4z + 109 – k2 = 0 2(x2 + y2 + z2) – 4x – 14y + 4z = k2 – 109 Dividing by 2 both sides (2(𝑥^2 + 𝑦^2 + 𝑧^2))/2−4𝑥/2−14𝑦/2+4𝑧/2=(𝑘^2 − 109)/2 x2 + y2 + z2 – 2x – 7y + 2z = (𝑘^2 − 109)/2 Thus the required equation is x2 + y2 + z2 – 2x – 7y + 2z = (𝒌^𝟐 − 𝟏𝟎𝟗)/𝟐

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.