Misc 6 - If A and B be points (3, 4, 5), (-1, 3, -7) - Distance between two points - Set of points

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  1. Chapter 12 Class 11 Introduction to Three Dimensional Geometry
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Misc 6 If A and B be the points (3, 4, 5) and (–1, 3, –7), respectively, find the equation of the set of points P such that PA2 + PB2 = k2, where k is a constant. Given points A (3, 4, 5) & B ( – 1, 3, –7) Let Point P be (x, y, z,) We need to point equation of points P, such that PA2 + PB2 = k2 Calculating PA2 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 = 3, y2 = 4, z2 = 5 PA = ﷐﷮﷐3−x﷯2+﷐4−y﷯2+﷐5−z﷯2﷯ Squaring both sides (PA)2 = ﷐﷮﷐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 – 6x – 8y – 10z + 9 + 16 + 25 = x2 + y2 + z2 – 6x – 8y – 10z + 50 Calculating PB2 P (x, y, z) , B ( – 1, 3 , – 7) PB = ﷐﷮﷐x2−x1﷯2+﷐y2−y1﷯2+﷐z2 −z1﷯2﷯ Here, x1 = x, y1 = y, z1 = z x2 = –1, y2 = 3, z2 = –7 PB = ﷐﷮﷐−1−x﷯2+﷐3−y﷯2+﷐−7−z﷯2﷯ PB = ﷐﷮﷐−1﷯2 ﷐1+𝑥﷯2+﷐3−y﷯2+﷐−1﷯2﷐7−z﷯2﷯ PB = ﷐﷮﷐1+𝑥﷯2+﷐3−y﷯2+﷐7−z﷯2﷯ Squaring both sides (PB)2 = ﷐﷐﷐﷮﷐1+𝑥﷯2+﷐3−y﷯2+﷐7+z﷯2﷯﷯﷮2﷯ (PB)2 = (1 + x)2 + (3 – y)2 + (7 + z)2 (PB)2 = (1)2 + x2 + 2(1)(x) + (3)2 + (y)2 – 2(3)(y) + (7)2 + z2 + 2(7)(z) = 1 + x2 + 2x + 9 + y2 – 6y + 49 + y2 + +14z = x2 + y2 + z2 + 2x – 6y + 14z + 1 + 9 + 49 = x2 + y2 + z2 + 2x – 6y + 14z + 59 Putting value of (PA)2 & (PB)2 in (1) (PA)2 + (PB)2 = k2 (x2 + y2 + z2 – 6x – 8y – 10z + 50) + (x2 + y2 + z2 + 2x – 6y + 14z + 59) = k2 2x2 + 2y2 + 2z2 – 6x + 2x – 8y – 6y – 10z + 4z + 50 + 59 = 52 2x2 + 2y2 + 2z2 – 4x – 14y – 6z + 109 = 52 2(x2 + y2 + z2) – 4x – 14y – 6z + 109 – 52 = 0 Thus the required equation is 2(x2 + y2 + z2) – 4x – 14y – 6z + 109 – 52 = 0

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