   1. Chapter 12 Class 11 Introduction to Three Dimensional Geometry
2. Serial order wise
3. Miscellaneous

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﷯ We know that Distance between two point (x1 y1 z1) & (x2 y2 z2) is D = ﷐﷮﷐x2−x1﷯2+﷐y2−y1﷯2+﷐z2 −z1﷯2﷯ 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+9+4+25﷯ 5﷐﷮2﷯ = ﷐﷮𝑎2+4𝑎+38﷯ (5﷐﷮2﷯)2 = ﷐﷐﷮𝑎2+4𝑎+38﷯﷯2 5 × 5 × ﷐﷮2﷯ × ﷐﷮2﷯ = 𝑎2+4𝑎+38 25 × 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)

Miscellaneous 