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Ex 12.2, 4 - Find equation of set of points equidistant - Ex 12.2

Ex 12.2,  4 - Chapter 12 Class 11 Introduction to Three Dimensional Geometry - Part 2
Ex 12.2,  4 - Chapter 12 Class 11 Introduction to Three Dimensional Geometry - Part 3 Ex 12.2,  4 - Chapter 12 Class 11 Introduction to Three Dimensional Geometry - Part 4 Ex 12.2,  4 - Chapter 12 Class 11 Introduction to Three Dimensional Geometry - Part 5

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Ex 12.2, 4 Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, –1). Let A (1, 2, 3) & B (3, 2, –1) Let point P be (x, y, z,) Let point P (x, y, z) be at equal distance from point A (1, 2, 3) & B (3, 2, – 1) i.e. PA = PB Calculating PA P (x, y, z), A (1, 2, 3) PA = √((x2−x1)2+(y2−y1)2+(z2 −z1)2) Here, x1 = x, y1 = y, z1 = z x2 = 1, y2 = 2, z2 = 1 PA = √((1−x)2+(2−y)2+(3 −z)2) Calculating PB P (x, y, z), B (3, 4, 5) PB = √((x2−x1)2+(y2−y1)2+(z2 −z1)2) Here, x1 = x, y1 = y, z1 = z x2 = 3, y2 = 2, z2 = −1 PB = √((3−x)2+(2−y)2+(−1 −z)2) PB = √((3−x)2+(2−y)2+(−1)2(1+𝑧)2) = √((3−x)2+(2−y)2+(1+𝑧)2) Since PA = PB √((1+𝑧)2+(2−y)2+(3−x)2) = √((3−x)2+(2−y)2+(1+𝑧)2) Squaring Both sides (√((1+𝑧)2+(2−y)2+(3−x)2))^2= (√((3−x)2+(2−y)2+(1+𝑧)2))^2 (1 – x)2 + (2 – y)2 + (3 – z)2 = (3 – x)2 + (2 – y)2 + (1 + z)2 (1)2 + (x)2 – 2(1)(x) + (4)2 + (y)2 – 2(2)(y) + (3)2 + (z)2 + 2(3)(z) = (3)2 + (x)2 – 2(3)(x) + (2)2 + (y)2 – 2(2)(y) + (1)2 + (z)2 + 2(1)(z) 1 + x2 – 2x + 42 + y2 – 4y + 9 + z2 – 6z = 9 + x2 – 6x + 4 + y2 – 4y + 1 + z2 + 2z – 2x – 4y – 6z + 14 = (x2 + y2 + z2) – (x2 + y2 – z2) – 6x – 4y + 2z + 14 – 2x – 4y – 6z + 14 = – 6x –4y + 2z + 14 – 2x – 4y – 6z = – 6x – 4y + 2z – 2x + 6x – 4y + 4y – 6z –2z = 0 4x + 0 – 8z = 0 4x – 8z = 0 4(x – 2z) = 0 x – 2z = 0 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.