# [SOLVED] Area of right triangle with 3 vertices

• Mar 6th 2010, 05:34 AM
thekrown
[SOLVED] Area of right triangle with 3 vertices
I want to find the area of a right triangle with vertices P,Q,R.

P (1,-1,2)
Q (0,3,4)
R (6,1,8)

Area of right triangle is |PQ x PR| / 2

vector PQ (-1,4,2)
vector PR (5,2,6)
cross product vector PQ and vector PR (20,-16,-22)

The area should be |20^2 + -16^2 + -22^2| / 2

= square root (20^2 + -16^2 + -22^2) / 2

= 570

I think I got this, I'm just doubting the formula. I have trouble knowing when |...| means what if it's absolute value or the square root.
• Mar 6th 2010, 05:58 AM
HallsofIvy
"| |" never means "square root". It means absolute value if it is around a number, length if it is around a vector. Of course, you then use the square root to find the length of the vector (and to find an absolute value: $|x|= \sqrt{x^2}$) but it is incorrect to say that it means square root!

By the way, this is NOT a right triangle but, fortunately, the area of any triangle, "right" or not, with two sides formed by vectors $\vec{u}$ and $\vec{v}$ is $\frac{1}{2}|\vec{u}\times\vec{v}|$. There the | | clearly means "length" (or "magnitude") because $\vec{u}\times\vec{v}$ is a vector.
• Mar 6th 2010, 06:04 AM
thekrown
I see, so the abs value is put there since an area really isn't negative. How are my calculations?