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Area of a Triangle - by Heron's Formula

Area of a rec...

Question

Area of a rectangle having vertices $A, B, C$ and $D$ with position vectors $-^i + \frac{1}{2}^j + 4^k,^i + \frac{1}{2}^j + 4^k,^i - \frac{1}{2}^j + 4^k and -^i - \frac{1}{2}^j + 4^k$ respectively is

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Solution

$- - \to A B = - - \to O B - - - \to O A = (^i + \frac{1}{2}^j + 4^k) - (-^i + \frac{1}{2}^j + 4^k) = 2^i - - \to A D = - - \to O D - - - \to O A = (-^i - \frac{1}{2}^j + 4^k) - (-^i + \frac{1}{2}^j + 4^k) = -^j$
Area of rectangle $= | - - \to A B \times - - \to A D | = | (2^i) \times (-^j) | = | - 2^k | = \sqrt{(- 2)^{2}} = 2$

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Area of a Triangle - by Heron's Formula

Standard IX Mathematics

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