# Ex 12.2, 4 - 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

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,) Since it is given that point P (x, y, z) is 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−x12+y2−y12+z2 −z12 Here, x1 = x, y1 = y, z1 = z x2 = 1, y2 = 2, z2 = 3 PA = 1−x2+2−y2+3 −z2 Calculating PB P (x, y, z) B (3, 4, 5) PB = x2−x12+y2−y12+z2 −z12 Here, x1 = x, y1 = y, z1 = z x2 = 3, y2 = 4, z2 = 5 PB = 3−x2+2−y2+−1 −z2 PB = 3−x2+2−y2+−121+𝑧2 = 3−x2+2−y2+1+𝑧2 Since PA = PB 1+𝑧2+2−y2+3−x2 = 3−x2+2−y2+1+𝑧2 Squaring Both sides 1+𝑧2+2−y2+3−x22= 3−x2+2−y2+1+𝑧22 (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 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.