Example, 11 - Chapter 12 Class 11 Introduction to Three Dimensional Geometry
Last updated at Jan. 31, 2020 by Teachoo
Last updated at Jan. 31, 2020 by Teachoo
Transcript
Example 11 Show that the points A (1, 2, 3), B (–1, –2, –1), C (2, 3, 2) & 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 Opposite sides are equal + Diagonals are equal Opposite sides are equal Thus, We need to prove that Opposite side are equal i.e. AB = CD and BC = DA but Diagonals Not equal ( AC ≠ BD) Calculating AB 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 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 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 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 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) = √(12+12+(−1)2) = √(1+1+1) = √3 Calculating BD 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
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