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Classification of Triangles Based on Angles

Let a⃗ , b⃗...

Question

Let $\to a, \to b, \to c$ be position vectors of three vertices of triangle $A B C,$ find the area of triangle $A B C .$

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Solution

$cos θ = \frac{- - \to A D}{A B} = \frac{∣ ∣ ∣ - - \to A D ∣ ∣ ∣}{∣ ∣ ∣ \to b - \to a ∣ ∣ ∣}$
$\Rightarrow ∣ ∣ ∣ - - \to A D ∣ ∣ ∣ =$ height $= ∣ ∣ ∣ \to b - \to a ∣ ∣ ∣ cos θ$
and $cos θ = \frac{(\to c - \to b) . (\to a - \to b)}{∣ ∣ ∣ \to c - \to b ∣ ∣ ∣ ∣ ∣ ∣ \to a - \to b ∣ ∣ ∣}$ (Using $cos θ = \frac{\to a . \to b}{∣ ∣ \to a ∣ ∣ ∣ ∣ ∣ \to b ∣ ∣ ∣}$ )
$∴$ height $= ∣ ∣ ∣ \to b - \to a ∣ ∣ ∣ ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ \frac{(\to c - \to b) . (\to a - \to b)}{∣ ∣ ∣ \to c - \to b ∣ ∣ ∣ ∣ ∣ ∣ \to a - \to b ∣ ∣ ∣} ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ = \frac{(\to c - \to b) . (\to a - \to b)}{∣ ∣ ∣ \to c - \to b ∣ ∣ ∣}$
$∴$ area of triangle $= \frac{1}{2} \times$ base height
$= \frac{1}{2} (\to c - \to b) ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ \frac{(\to c - \to b) . (\to a - \to b)}{∣ ∣ ∣ \to c - \to b ∣ ∣ ∣} ⎫ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎭$

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