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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. |
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Answer» With the given vertices, `AB = sqrt((-2)^2+(-4)^2+(-4)^2 ) = sqrt(4+16+16) = sqrt 36 = 6` `BC = sqrt((3)^2+(5)^2+(3)^2 ) = sqrt(9+25+9) = sqrt 43 ` `CD = sqrt((2)^2+(4)^2+(4)^2 ) = sqrt(4+16+16) = sqrt 36 = 6` `AD = sqrt((3)^2+(5)^2+(3)^2 ) = sqrt(9+25+9) = sqrt 43 ` Here, `AB = CD` and `BC = AD`. So, `ABCD` can be a parallelogram or a rectangle. For `ABCD` to be a rectangle, all angles should be `90^@`. So, it should satisfy pythagoras theorem. Now, `BD = sqrt((5)^2+(9)^2+(7)^2 ) = sqrt(25+81+49) = sqrt 155` Now, `BC^2 + CD^2 = 43+36 = 79` `BD^2 = 155` As, `BD^2 != BC^2 + CD^2` So, `Delta BCD` is not a right angle triangle.Thus, `ABCD` is a parallelogram but not a rectangle. |
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