Misc 4 - Find coordinates of a point on y-axis - Chapter 12

Misc  4 - Chapter 12 Class 11 Introduction to Three Dimensional Geometry - Part 2
Misc  4 - Chapter 12 Class 11 Introduction to Three Dimensional Geometry - Part 3

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Transcript

Misc 4 Find the coordinates of a point on y-axis which are at a distance of 5√2 from the point P (3, –2, 5). Since point is on y-axis its x & z coordinate is 0 Let the point on y-axis be A(0, a, 0) Given that Point A is at a distance of 5√2 from point P (3, – 2, 5) i.e. PA = 5√2 Finding Distance PA Calculating PA P (3, –2, 5) , A (0, a, 0) Here x1 = 3, y1 = – 2, z1 = 5 x2 = 0, y2 = a, z2 = 0 PA = √((0−3)2+(a−(−2))2+(0 −5)2) 5√2 = √((−3)2+(a+2)2+(0−5)2) 5√2 = √(9+(𝑎+2)2+25) 5√2 = √(9+𝑎2+(2)2+2(𝑎)(2)+25) 5√2 = √(9+𝑎2+4+4𝑎+25) 5√2 = √(𝑎2+4𝑎+38) Squaring both sides (5√2)2 = (√(𝑎2+4𝑎+38))2 5 × 5 × √2 × √2 = 𝑎2+4𝑎+38 50 = 𝑎2+4𝑎+38 𝑎2+4𝑎+38 = 50 𝑎2+4𝑎+38 – 50 = 0 𝑎2+4𝑎− 12 = 0 𝑎2+6𝑎− 2a – 12 = 0 a(a + 6) – 2 (a + 6) = 0 (a – 2) (a + 6 ) = 0 So, a = 2 or a = –6 Thus, the coordinates of the required point is (0, 2, 0) & (0, –6, 0)

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.