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I have three points with coordinates: $A (5,-1,0),B(2,4,10)$, and $C(6,-1,4)$.

I have the following vectors $\overrightarrow {CA} = (-1, 0, -4)$ and $\overrightarrow{CB} = (-4, 5, 6)$.

To find the area of the triangle I used the dot product between these vectors to get the angle and then applied the formula $A=0.5ab\sin{C}$ to find the area of the triangle which gave me $15.07(2dp)$.

However in the given solutions the answer is given as $(3*\sqrt(102))/2$

I think they have used the trig identity $\cos^2(\theta) + \sin^2(\theta) = 1$ to find the value of $\sin(\theta)$ rather than $\arccos(\theta)$ to find the angle ACB. However I don't understand why there would be such a discrepancy between the two answers; one using $\arccos$ and the other using the trig identity.

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  • $\begingroup$ Recommendation to a newbie: Do not use one symbol for two items, e.g., $A$ as point on triangle and $A$ as area of the triangle. $\endgroup$ Feb 16 at 23:47
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Note that $a=\sqrt{17}$, $b=\sqrt{77}$ and $\cos C= -\frac{20}{ab}$, which yields the area

$$A= \frac12 ab \sqrt{1-\cos^2C}=\frac12 \sqrt{17\cdot77-400}=\frac{3\sqrt{101}}2$$

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If you calculate the angle, you introduce numerical errors. The most elegant way to calculate the area is to use the cross product. You will not need any trigonometric functions: $$A=\frac12|\vec{CA}\times\vec{CB}|$$ So all you need will be some multiplications, additions(subtractions), and a square root. Let me know if you can follow this path. By the way, the answer should be $\frac32\sqrt{101}$, if I did the math right, which is closer to your calculations

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Your book got a typo: it's $101$ and not $102$. And that's it, that is approximately your answer.

Hope this helps. Ask anything if not clear :)

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