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

Transcript

Ex 12.2, 1 Find the distance between the following pairs of points: (i) (2, 3, 5) and (4, 3, 1) Let P be (2, 3, 5) and Q be (4, 3, 1) Distance PQ = ﷐﷮﷐x2−x1﷯2+﷐y2−y1﷯2+﷐z2 −z1﷯2﷯ Here, x1 = 2, y1 = 3, z1 = 5 x2 = 4, y2 = 3, z2 = 1 PQ = ﷐﷮﷐4−2﷯2+﷐3−3﷯2+﷐1 −5﷯2﷯ = ﷐﷮22+02+﷐−4﷯2﷯ = ﷐﷮4+0+16﷯ = ﷐﷮20﷯ = ﷐﷮2 ×2 ×5﷯ = 2﷐﷮5﷯ Thus, the required distance is 2﷐﷮5﷯ units Ex12.2, 1 Find the distance between the following pairs of points: (ii) (–3, 7, 2) and (2, 4, –1) Let Point P be ( –3, 7, 2) Q be (2, 4, –1) PQ = ﷐﷮﷐x2−x1﷯2+﷐y2−y1﷯2+﷐z2 −z1﷯2﷯ Here x1 = –3, y1 = 7, z1 = 2 x2 = 2, y2 = 4, z2 = – 1 PQ = ﷐﷮﷐2−(−3)﷯2+﷐4−7﷯2+﷐−1 −2﷯2﷯ = ﷐﷮﷐2+3﷯2+﷐−3﷯2+﷐−3﷯2﷯ = ﷐﷮﷐5﷯2+9+9﷯ = ﷐﷮25+9+9﷯ = ﷐﷮43﷯ Thus, the required distance is ﷐﷮43﷯ units Ex12.2, 1 Find the distance between the following pairs of points: (iii) (–1, 3, –4) and (1, –3, 4) Let Point P be ( –1, 3, –4) Q be (1, –3, 4) PQ = ﷐﷮﷐x2−x1﷯2+﷐y2−y1﷯2+﷐z2 −z1﷯2﷯ x1 = –1, y1 = 3, z1 = -4 x2 = 1, y2 = – 3, z2 = 4 PQ = ﷐﷮﷐1−(−1﷯2+﷐−3−3﷯2+﷐4−(−4﷯2﷯ = ﷐﷮﷐1+1﷯2+﷐−6﷯2+﷐4+4﷯2﷯ = ﷐﷮﷐2﷯2+36+﷐8﷯2﷯ = ﷐﷮4+36+64﷯ = ﷐﷮104﷯ = ﷐﷮26×2 × 2﷯ = 2 ﷐﷮26﷯ Thus, the required distance is 2﷐﷮26﷯ units Ex12.2, 1 Find the distance between the following pairs of points: (iv) (2, –1, 3) and (–2, 1, 3) Let Point P be ( 2, –1, 3) Q be (– 2, 1, 3) PQ = ﷐﷮﷐x2−x1﷯2+﷐y2−y1﷯2+﷐z2 −z1﷯2﷯ x1 = 2, y1 = –1, z1 = 3 x2 = – 2, y2 = 1, z2 = 3 PQ = ﷐﷮﷐−2−2﷯2+﷐1−(−1﷯2+﷐3−3﷯2﷯ = ﷐﷮﷐−4﷯2+﷐1+1﷯2+﷐0﷯2﷯ = ﷐﷮16+﷐2﷯2+02﷯ = ﷐﷮16+4+0﷯ = ﷐﷮20﷯ = ﷐﷮5×2×2﷯ = 2﷐﷮5﷯ Thus, the required distance is 2﷐﷮5﷯ units

Ex 12.2 