Find the area of the triangle with the following vertices: $(a, 0, 0), (0, b, 0), (0, 0, c) .$

\frac{\sqrt{(a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2})}}{2}

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(a b c) / 2

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\frac{1}{2} (a^{2} + b^{2} + c^{2})

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none of these

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Solution

The correct option is B $\frac{\sqrt{(a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2})}}{2}$
Let $A (a, 0, 0), B (0, b, 0)$ and $C (0, 0, c$
Then $- - \to A B = B - A = (0, b, 0) - (a, 0, 0) = (- a, b, 0)$
And $- - \to A C = C - A = (0, 0, c) - (a, 0, 0) = (- a, 0, c)$
Now $- - \to A B \times - - \to A C = ∣ ∣ ∣ ∣ \begin{matrix} i & j & k - a & b & 0 - a & 0 & c \end{matrix} ∣ ∣ ∣ ∣ = i (b c - 0) - j (- a c - 0) + k (0 + a b) = b c i + a c j + a b k$
Therefore area of $Δ A B C = \frac{∣ ∣ ∣ - - \to A B \times - - \to A C ∣ ∣ ∣}{2} = \frac{\sqrt{{(b c)}^{2} + {(a c)}^{2} + {(a b)}^{2}}}{2}$

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