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Where $a = ( 1, 2, 3), b = (2, 1, 3) $, and $c = (3,1,2). $

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  • $\begingroup$ We can find the lengths of the sides and apply Heron's Formula. $\endgroup$ – Indrayudh Roy Apr 3 '14 at 14:26
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Heron works of course but it would be simpler to take half the length of the cross product $(b-a)\times(c-a)$.

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Solution: Construct the vectors $\hat{ab}$, $\hat{ac}$ and take $\frac{1}{2} |\hat{ab} \times \hat{ac}|$. We take half of the resulting since the original gives the area of the parallelogram decsribed by the vectors.

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use this formula: $$S=\sqrt{p(p-s_1)(p-s_2)(p-s_3)}$$ where $p=\frac{s_1+s_2+s_3}{2}$, and $s_1,s_2,s_3$ are the lengths of the three sides. You can get $s_1,s_2,s_3$ from the three positions.

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Compute the differences $a-b$ and $a-c$. then take the cross product of them $(a-c)\times(a-c)$. The norm of this, divided by two is the area of the triangle.

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One way to do it would be to first find out the length of each of the 3 sides, by simply applying Pythagorus. e.g.

AB = $\sqrt{(2-0)^2 + (1-1)^2 + (4-3)^2}=\sqrt5$

You can do this for all of the sides, and it is possible to find the area of a triangle given the length of each of its sides.

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Cross product works great as a black box, but it lacks geometric intuition.

enter image description here

For triangle ABC, we construct a line segment intersecting C, and normal to side AB. The triangle's area is $0.5\cdot \left \| \overrightarrow{AB} \right \|\cdot \left \| \overrightarrow{CH} \right \|$ $$\left \| \overrightarrow{CH} \right \|=\left \| \overrightarrow{AC} \right \| \cdot sin(θ)$$ where $$θ = arccos(\frac{\overrightarrow{AB} \cdot \overrightarrow{AC}}{\left \| \overrightarrow{AB} \right \| \cdot \left \| \overrightarrow{AC} \right \|})$$

Thus: $$Area=0.5\cdot \left \| \overrightarrow{AB} \right \|\cdot \left \| \overrightarrow{AC} \right \|\cdot sin(θ)$$ This is equal to: $$0.5\cdot \left \| \overrightarrow{AB} \times \overrightarrow{AC} \right \|$$

This is why $$0.5\cdot \left \| \overrightarrow{AB} \times \overrightarrow{AC} \right \|=0.5\cdot \left \| \overrightarrow{AB} \right \|\cdot \left \| \overrightarrow{AC} \right \|\cdot sin(θ)$$

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