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.