The area of the quadrilateral ABCD, where A(0,4,1), B(2, 3, –1), C(4, 5, 0) and D(2, 6, 2), is equal to ______.

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The area of the quadrilateral ABCD, where A(0,4,1), B(2, 3, –1), C(4, 5, 0) and D(2, 6, 2), is equal to 9 sq.units. Explanation: Given points are A(0, 4, 1), B(2,3,– 1), C(4, 5, 0) and D(2,6,2) D’ratios of AB = 2,–1 –2 And d’ratios of DC = 2,–1,–2 $\therefore$ AB||DC Similarly, d’ratios of AD = 2, 2, 1 and d’ratios of BC = 2, 2, 1 $\therefore$ AD || BC So ABCD is a parallelogram `vec"AB" = 2hat"i" - hat"j" - 2hat"k" vec"AD" = 2hat"i" + 2hat"j" + hat"k"` $\therefore$ Area of parallelogram ABCD = |vec"AB" xx vec"AD"| = `|(hat"i", hat"j", hat"k"),(2, -1, -2),(2, 2, 1)| = hat"i"(-1 + 4) - hat"j"(2 + 4) + hat"k"(4 + 2) = 3hat"i" - 6hat"j" + 6hat"k" = sqrt((3)^2 + (-6)^2 + (6)^2) = sqrt(9 + 36 + 36) = sqrt(81)` = 9 sq.units [Solution diagram: 254167_solution_1.png]

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