    1. Chapter 12 Class 11 Introduction to Three Dimensional Geometry
2. Serial order wise
3. Ex 12.2

Transcript

Ex 12.2, 4 Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, –1). Let A (1, 2, 3) & B (3, –2, –1) Let point P be (x, y, z,) Since it is given that point P (x, y, z) is equal distance from point A (1, 2, 3) & B (3, 2, – 1) i.e. PA = PB Calculating PA P (x, y, z) A (1, 2, 3) PA = ﷐﷮﷐x2−x1﷯2+﷐y2−y1﷯2+﷐z2 −z1﷯2﷯ Here, x1 = x, y1 = y, z1 = z x2 = 1, y2 = 2, z2 = 3 PA = ﷐﷮﷐1−x﷯2+﷐2−y﷯2+﷐3 −z﷯2﷯ Calculating PB P (x, y, z) B (3, 4, 5) PB = ﷐﷮﷐x2−x1﷯2+﷐y2−y1﷯2+﷐z2 −z1﷯2﷯ Here, x1 = x, y1 = y, z1 = z x2 = 3, y2 = 4, z2 = 5 PB = ﷐﷮﷐3−x﷯2+﷐2−y﷯2+﷐−1 −z﷯2﷯ PB = ﷐﷮﷐3−x﷯2+﷐2−y﷯2+﷐−1﷯2﷐1+𝑧﷯2﷯ = ﷐﷮﷐3−x﷯2+﷐2−y﷯2+﷐1+𝑧﷯2﷯ Since PA = PB ﷐﷮﷐1+𝑧﷯2+﷐2−y﷯2+﷐3−x﷯2﷯ = ﷐﷮﷐3−x﷯2+﷐2−y﷯2+﷐1+𝑧﷯2﷯ Squaring Both sides ﷐﷐﷐﷮﷐1+𝑧﷯2+﷐2−y﷯2+﷐3−x﷯2﷯﷯﷮2﷯= ﷐﷐﷐﷮﷐3−x﷯2+﷐2−y﷯2+﷐1+𝑧﷯2﷯﷯﷮2﷯ (1 – x)2 + (2 – y)2 + (3 – z)2 = (3 – x)2 + (2 – y)2 + (1 + z)2 (1)2 + (x)2 – 2(1)(x) + (4)2 + (y)2 – 2(2)(y) + (3)2 + (z)2 – 2(3)(z) = (3)2 + (x)2 – 2(3)(x) + (2)2 + (y)2 – 2(2)(y) + (1)2 + (z)2 – 2(1)(z) 1 + x2 – 2x + 42 + y2 – 4y + 9 + z2 – 6z = 9 + x2 – 6x + 4 + y2 – 4y + 1 + z2 + 2z – 2x – 4y – 6z + 14 = (x2 + y2 + z2) – (x2 + y2 – z2) – 6x – 4y + 2z + 14 – 2x – 4y – 6z + 14 = – 6x –4y + 2z + 14 – 2x – 4y – 6z = – 6x – 4y + 2z – 2x + 6x – 4y + 4y – 6z –2z = 0 4x + 0 – 8z = 0 4x – 8z = 0 4(x – 2z) = 0 x – 2z = 0 which is the required equation

Ex 12.2 