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

Transcript

Example, 11 Show that the points A (1, 2, 3), B (–1, –2, –1), C (2, 3, 2) and D (4, 7, 6) are the vertices of a parallelogram ABCD, but it is not a rectangle. Difference between We need to prove that ABCD is a parallelogram but not a rectangle So, We need to prove that Opposite side are equal i.e. AB = CD and BC = DA but Diagonals Not equal ( AC ≠ BD) Calculating AB A (1, 2, 3) B ( – 1, – 2, – 1) AB = ﷐﷮﷐x2−x1﷯2+﷐y2−y1﷯2+﷐z2 −z1﷯2﷯ x1 = 1, y1 = 2, z1 = 3 x2 = – 1, y2 = – 2, z2 = – 1 AB = ﷐﷮﷐−1−1﷯2+﷐−2−2﷯2+﷐−1−3﷯2﷯ = ﷐﷮﷐−2﷯2+﷐−4﷯2+﷐−4﷯2﷯ = ﷐﷮4+16+16﷯ = ﷐﷮36﷯ = 6 Calculating BC B ( – 1, – 2, – 1) C (2, 3, 2) BC = ﷐﷮﷐x2−x1﷯2+﷐y2−y1﷯2+﷐z2 −z1﷯2﷯ x1 = – 1, y1 = – 2, z1 = – 1 x2 = 2, y2 = 3, z2 = 2 BC = ﷐﷮﷐2−(−1﷯2+﷐3−(−2)﷯2+﷐2−(−1)﷯2﷯ = ﷐﷮﷐2+1﷯2+﷐3+2﷯2+﷐2+1﷯2﷯ = ﷐﷮﷐3﷯2+﷐5﷯2+﷐3﷯2﷯ = ﷐﷮9+25+9﷯ = ﷐﷮43﷯ Calculating CD C (2, 3, 2) D (4, 7, 6) CD = ﷐﷮﷐x2−x1﷯2+﷐y2−y1﷯2+﷐z2 −z1﷯2﷯ x2 = 2, y2 = 3, z2 = 2 x2 = 4, y2 = 7, z2 = 6 CD = ﷐﷮﷐4−2﷯2+﷐7−3﷯2+﷐6−2﷯2﷯ = ﷐﷮﷐2﷯2+﷐4﷯2+﷐4﷯2﷯ = ﷐﷮4+16+16﷯ = ﷐﷮36﷯ = 6 Calculating DA D (4, 7, 6) A (1, 2, 3) DA = ﷐﷮﷐x2−x1﷯2+﷐y2−y1﷯2+﷐z2 −z1﷯2﷯ x2 = 4, y2 = 7, z2 = 6 x1 = 1, y1 = 2, z1 = 3 DA = ﷐﷮﷐1−4﷯2+﷐2−7﷯2+﷐3−6﷯2﷯ = ﷐﷮﷐−3﷯2+﷐−4﷯2+﷐−3﷯2﷯ = ﷐﷮9+16+9﷯ = ﷐﷮43﷯ Hence AB = CD = 6 & BC = DA = ﷐﷮43﷯ Thus, Opposite side are equal ∴ ABCD is Parallelogram Now we need to prove, diagonals are not equal i.e. AC ≠ BD Calculating AC A (1, 2, 3) C (2, 3, 2) AC = ﷐﷮﷐x2−x1﷯2+﷐y2−y1﷯2+﷐z2 −z1﷯2﷯ Here, x2 = 1, y2 = 2, z2 = 3 x1 = 2, y1 = 3, z1 = 2 AC = ﷐﷮﷐2−1﷯2+﷐3−2﷯2+﷐2−3﷯2﷯ = ﷐﷮﷐1﷯2+﷐1﷯2+﷐−1﷯2﷯ = ﷐﷮1+1+1﷯ = ﷐﷮3﷯ Calculating BD B ( – 1, – 2, – 1) , D (4, 7, 6) BD = ﷐﷮﷐x2−x1﷯2+﷐y2−y1﷯2+﷐z2 −z1﷯2﷯ Here, x1 = –1, y1 = –2, z1 = –1 x2 = 4, y2 = 7, z2 = 6 BD = ﷐﷮﷐x2−x1﷯2+﷐y2−y1﷯2+﷐z2 −z1﷯2﷯ BD = ﷐﷮﷐4−(−1)﷯2+﷐7−(−2)﷯2+﷐6−(−1)﷯2﷯ = ﷐﷮﷐4+1﷯2+﷐7+2﷯2+﷐6+1﷯2﷯ = ﷐﷮﷐5﷯2+﷐9﷯2+﷐7﷯2﷯ = ﷐﷮25+81+49﷯ = ﷐﷮155﷯ Now, AC = ﷐﷮3﷯ & BD = ﷐﷮155﷯ As ﷐﷮3﷯ ≠ ﷐﷮155﷯ Hence AC ≠ BD Since Diagonals Not equal Hence it is ABCD is not rectangle ∴ ABCD is a parallelogram but not a rectangle Hence proved

Examples 