# Example 12 - Chapter 12 Class 11 Introduction to Three Dimensional Geometry

Last updated at Jan. 11, 2019 by Teachoo

Last updated at Jan. 11, 2019 by Teachoo

Transcript

Example 12 Find the equation of the set of the points P such that its distances from the points A (3, 4, –5) and B (– 2, 1, 4) are equal. Given A (3, 4, 5) & B ( –2, 1, 4,) Let point P be (x, y, z,) Given PA = PB Calculating PA 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 = – 2, y2 = 1, z2 = 4 PA = √((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+𝑧2−6𝑥−8𝑦+10𝑧 9+16+25) = √(𝑥2+𝑦2+𝑧2−6𝑥−8𝑦+10𝑧+40) Calculating PB P (x, y, z) B ( –2, 1, 4) PB = √((x2−x1)2+(y2−y1)2+(z2 −z1)2) Here, x1 = x, y1 = y, z1 = z x2 = – 2, y2 = 1, z2 = 4 PB = √((−2−𝑥)2+(1−𝑦)2+(4−𝑧)2) = √((2+𝑥)2+(1−𝑦)2+(4−𝑧)2) = √((2)2+(𝑥)2+2(2)(𝑥)+(1)2+𝑦2−2(1)(𝑦)+42+𝑧2−2(4)(𝑧) ) = √(4+𝑥2+4𝑥+1+𝑦2−2𝑦+16+𝑧2−8𝑧) = √(𝑥2+𝑦2+𝑧2−4𝑥−2𝑦+8𝑧+21) = √(𝑥2+𝑦2+𝑧2−6𝑥−8𝑦+10𝑧+40) Now ,given that PA = PB √(𝑥2+𝑦2+𝑧2−6𝑥−8𝑦+10𝑧+40) = √(𝑥2+𝑦2+𝑧2+4𝑥−2𝑦+8𝑧+21) Squaring both sides (√(𝑥2+𝑦2+𝑧2−6𝑥−8𝑦+10𝑧+40))2 = (√(𝑥2+𝑦2+𝑧2+4𝑥−2𝑦+8𝑧+21))2 𝑥2+𝑦2+𝑧2−6𝑥−8𝑦+10𝑧+40 = 𝑥2+𝑦2+𝑧2+4𝑥−2𝑦+8𝑧+21 𝑥2+𝑦2+𝑧2−6𝑥−8𝑦+10𝑧+40 – 𝑥2−𝑦2−𝑧2+4𝑥+2𝑦+8𝑧−21=0 𝑥2−𝑥2+𝑦2+𝑦2+𝑧2−𝑧2−6𝑥−4𝑥+8𝑦+2y+10z+8z+40−21=0 0 + 0 + 0 – 10x – 6y + 18z + 29 = 0 –10x – 6y + 18z + 29 = 0 0 = 10x + 6y – 18z – 29 = 0 10x + 6y – 18z – 29 = 0 which is the required equation

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.