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When I looked at this problem I didn't think it seemed all that hard until I actually tried it. The problem is this:

Given the rectangular vertices $O(0, 0, 0), P(-1, 2, -3), Q(-2, 3, -4), R(0, 0, 1)$, find the area of the triangle formed.

So since I am given 4 points, I know this is a parallelepiped of some sort. The formula for the Area of a parallelogram is $A = | b \times c |$ so that means the Area of the triangle would be $A = |b \times c| / 2$ since triangle is one-half base(height).

So I know I need vectors in order to use these formulas so I get these vectors by subtracting two points. So $\overrightarrow{OP} = \langle -1-0, 2-0, -3-0 \rangle = \langle -1, 2, -3\rangle$. Likewise, $\overrightarrow{OQ} = \langle -2, 3, -4 \rangle$ and $\overrightarrow{OR} = \langle 0, 0, 1 \rangle$

Let's set $\mathbf{b} = \overrightarrow{OQ}$ and $\mathbf{c} = \overrightarrow{OR}$. Then we must find $|\mathbf{b} \times \mathbf{c}|$ and then times that by half to get the area of the triangle.

$\langle -2, 3, -4 \rangle \times \langle 0, 0, 1 \rangle = (3-0)i-(-2-0)j+(0-0)k = 3i+2j$

$|3i+2j|/2 = \sqrt{9+4}/2 = \sqrt{13}/2$

I did the cross product by hand and after multiplying by 1/2 I got the answer to be $\sqrt{13}/2$ but when I look at the answer key, the answer the professor wrote was $\sqrt{14}/2$.

I am wondering why I am 1 digit off in that upper radical. Can anyone help explain why I am getting 1 off, or maybe there is a mistake on the answer sheet? Thanks.

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  • $\begingroup$ Maybe if you show us what did you do by hand we may be able to help you. There is a possibility that you made a mistake in the cross product calculation. $\endgroup$ – Joaquin Liniado Jun 7 '15 at 19:08
  • $\begingroup$ Did the problem asked to find the area of the triangle? Why is there 4 points? Did the problem originally ask to find the volume of the parallelpiped formed by the four vertices? $\endgroup$ – MathNewbie Jun 7 '15 at 19:12
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    $\begingroup$ These four points are definitely not coplanar. So using another set of three points, I find that the area of the triangle is not $\sqrt{14}/2$. The question is which three coordinates or perhaps an error on the answer key? $\endgroup$ – MathNewbie Jun 7 '15 at 19:19
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    $\begingroup$ Check triangle PQR, that should give you $\sqrt{14}/2$. It would be beneficial if it stated which three points, but now you know. $\endgroup$ – MathNewbie Jun 7 '15 at 19:26
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    $\begingroup$ In my opinion, this had nothing to do with $O$ being an origin. But was a poorly worded question if your professor expects you to get one of the four correctly without specifying which specific triangle. $\endgroup$ – MathNewbie Jun 7 '15 at 19:33

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