# Ex 12.2, 1 - Chapter 12 Class 11 Introduction to Three Dimensional Geometry

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

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−x12+y2−y12+z2 −z12 Here, x1 = 2, y1 = 3, z1 = 5 x2 = 4, y2 = 3, z2 = 1 PQ = 4−22+3−32+1 −52 = 22+02+−42 = 4+0+16 = 20 = 2 ×2 ×5 = 25 Thus, the required distance is 25 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−x12+y2−y12+z2 −z12 Here x1 = –3, y1 = 7, z1 = 2 x2 = 2, y2 = 4, z2 = – 1 PQ = 2−(−3)2+4−72+−1 −22 = 2+32+−32+−32 = 52+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−x12+y2−y12+z2 −z12 x1 = –1, y1 = 3, z1 = -4 x2 = 1, y2 = – 3, z2 = 4 PQ = 1−(−12+−3−32+4−(−42 = 1+12+−62+4+42 = 22+36+82 = 4+36+64 = 104 = 26×2 × 2 = 2 26 Thus, the required distance is 226 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−x12+y2−y12+z2 −z12 x1 = 2, y1 = –1, z1 = 3 x2 = – 2, y2 = 1, z2 = 3 PQ = −2−22+1−(−12+3−32 = −42+1+12+02 = 16+22+02 = 16+4+0 = 20 = 5×2×2 = 25 Thus, the required distance is 25 units

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.