# Vectors in 3D and area of a triangle

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.

• 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. Feb 16 at 23:47

Note that $$a=\sqrt{17}$$, $$b=\sqrt{77}$$ and $$\cos C= -\frac{20}{ab}$$, which yields the area
$$A= \frac12 ab \sqrt{1-\cos^2C}=\frac12 \sqrt{17\cdot77-400}=\frac{3\sqrt{101}}2$$
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
Your book got a typo: it's $$101$$ and not $$102$$. And that's it, that is approximately your answer.