# Example, 11

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

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−x12+y2−y12+z2 −z12 x1 = 1, y1 = 2, z1 = 3 x2 = – 1, y2 = – 2, z2 = – 1 AB = −1−12+−2−22+−1−32 = −22+−42+−42 = 4+16+16 = 36 = 6 Calculating BC B ( – 1, – 2, – 1) C (2, 3, 2) BC = x2−x12+y2−y12+z2 −z12 x1 = – 1, y1 = – 2, z1 = – 1 x2 = 2, y2 = 3, z2 = 2 BC = 2−(−12+3−(−2)2+2−(−1)2 = 2+12+3+22+2+12 = 32+52+32 = 9+25+9 = 43 Calculating CD C (2, 3, 2) D (4, 7, 6) CD = x2−x12+y2−y12+z2 −z12 x2 = 2, y2 = 3, z2 = 2 x2 = 4, y2 = 7, z2 = 6 CD = 4−22+7−32+6−22 = 22+42+42 = 4+16+16 = 36 = 6 Calculating DA D (4, 7, 6) A (1, 2, 3) DA = x2−x12+y2−y12+z2 −z12 x2 = 4, y2 = 7, z2 = 6 x1 = 1, y1 = 2, z1 = 3 DA = 1−42+2−72+3−62 = −32+−42+−32 = 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−x12+y2−y12+z2 −z12 Here, x2 = 1, y2 = 2, z2 = 3 x1 = 2, y1 = 3, z1 = 2 AC = 2−12+3−22+2−32 = 12+12+−12 = 1+1+1 = 3 Calculating BD B ( – 1, – 2, – 1) , D (4, 7, 6) BD = x2−x12+y2−y12+z2 −z12 Here, x1 = –1, y1 = –2, z1 = –1 x2 = 4, y2 = 7, z2 = 6 BD = x2−x12+y2−y12+z2 −z12 BD = 4−(−1)2+7−(−2)2+6−(−1)2 = 4+12+7+22+6+12 = 52+92+72 = 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

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.