$A = (0, 1, 2), B = (3, 0, 1), C = (4, 3, 6), D = (2, 3, 2)$ are the rectangular cartesian co-ordinates. Find the area of the triangle ABC.

\sqrt{90},

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\sqrt{\frac{45}{2}}

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45

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18

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Solution

The correct option is B $\sqrt{90},$
Given, $A = (0, 1, 2), B = (3, 0, 1), C = (4, 3, 6), D = (2, 3, 2)$

$\Rightarrow$ $\to A B = (3 - 0) i + (0 - 1) j + (1 - 2) k$
$\Rightarrow$ $\to A B = 3 i - 1 j - 1 k$

$\Rightarrow$ $\to A C = (4 - 0) i + (3 - 1) j + (6 - 2) k$
$\Rightarrow$ $\to A C = 4 i + 2 j + 4 k$

$\Rightarrow$ Area of triangle ABC is $= \frac{1}{2} | \to A B \times \to A C |$
$\Rightarrow$ $\to A B \times \to A C = ∣ ∣ ∣ ∣ \begin{matrix} i & j & k 3 & - 1 & - 1 4 & 2 & 4 \end{matrix} ∣ ∣ ∣ ∣ = - 2 i - 16 j + 10 k = 2 (- i - 8 j + 5 k)$
$\Rightarrow$ Required area is $= \frac{1}{2} | \to A B \times \to A C | = \sqrt{90}$

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^{3})

△ A B C

cos A cos B cos C = \frac{\sqrt{3} - 1}{8}

sin A sin B sin C = \frac{3 + \sqrt{3}}{8}

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