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

Last updated at March 9, 2017 by Teachoo

Last updated at March 9, 2017 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−x12+y2−y12+z2 −z12 Here, x1 = x, y1 = y, z1 = z x2 = – 2, y2 = 1, z2 = 4 PA = 3−𝑥2+4−𝑦2+−5−𝑍2 = 3−𝑥2+4−𝑦2+5+𝑍2 = 32+𝑥2−23𝑥+42+𝑦2−2×4−𝑦+52+𝑧2+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−x12+y2−y12+z2 −z12 Here, x1 = x, y1 = y, z1 = z x2 = – 2, y2 = 1, z2 = 4 PB = −2−𝑥2+1−𝑦2+4−𝑧2 = 2+𝑥2+1−𝑦2+4−𝑧2 = 22+𝑥2+22𝑥+12+𝑦2−21𝑦+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+402 = 𝑥2+𝑦2+𝑧2+4𝑥−2𝑦+8𝑧+212 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 + 10z + 29 = 0 0 = 10x + 6y – 10z – 29 = 0 10x + 6y – 10z – 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 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.