# Thread: [SOLVED] is this the are of this triangle

1. ## [SOLVED] is this the are of this triangle

given vertices

O (0,0,0)
P (1,2,-2)
Q (0,2,-1)
R (1,0,1)

PQ (-1,0,1)
Coord xyz of Q minus coord xyz of P

PR (0,-2,3)
Coord xyz of R minus coord xyz of P

Edit: and then calc the cross product of pq x pr before applying the formula below
Then apply 1/2 |square root a^2 + b^2 + c^2| = square root of 17 / 2

2. Originally Posted by thekrown
given vertices

O (0,0,0)
P (1,2,-2)
Q (0,2,-1)
R (1,0,1)
You have four points and a triangle has only three vertices. Did you mean "quadrilateral"?

PQ (-1,0,1)
Coord xyz of Q minus coord xyz of P

PR (0,-2,3)
Coord xyz of R minus coord xyz of P

Then apply 1/2 |square root a^2 + b^2 + c^2| = square root of 17 / 2
You have completely ignored (0, 0, 0). Please state the problem as it really is!

3. The problem states I must use vertices P, Q and R to find the area of the triangle.

4. It would have been nice if you had told us that! We had no way of knowing what triangle you were talking about!

Yes, the PQ and PR vectors are exactly what you say and the area of the triangle is 1/2 the absolute value of the cross product of those vectors.

5. Is what you said the same as the formula I used on the cross product?

ex: I put the cross product (2,3,2) where a=2, b=3, c=2 into the formula: 1/2 |square root a^2+b^2+c^2|

6. okay so the cross product is (2,3,2)

does this mean the area is 1/2 (half) the absolute value of 7?

7. Originally Posted by thekrown
okay so the cross product is (2,3,2)

does this mean the area is 1/2 (half) the absolute value of 7?
Then length of vector (2, 3, 2) is $\displaystyle \sqrt{4+ 9+ 4}= \sqrt{17}$, not 7. That gives the result you had initially.