Example 8 - Chapter 12 Class 11 Introduction to Three Dimensional Geometry (Important Question)
Last updated at May 6, 2024 by Teachoo
Chapter 12 Class 11 Introduction to Three Dimensional Geometry
Chapter 12 Class 11 Introduction to Three Dimensional Geometry
Last updated at May 6, 2024 by Teachoo
Example 8 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 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) = √((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+50) 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) 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