# Thread: Area of Triangle using cross product

1. ## Area of Triangle using cross product

A triangle has vertices A(-2,1,3), B(7,8,-4), and C(5,0,2). Determine the area of triangle ABC.

Vector AB = B-A = [9,7,-7]
Vector AC = C-A = [7, -1, -1]

Area of triangle = (ACxAB)/2

I end up getting vector [-14, -40, -40] and the magnitude is not the correct answer of 35.9 square units. Can someone help?

2. $\displaystyle AB \times AC=(-14,-40,-58)$

3. Hello, millerst!

A small error . . .

A triangle has vertices A(-2,1,3), B(7,8,-4), and C(5,0,2).
Determine the area of triangle ABC.

$\displaystyle \overrightarrow{AB} \:=\:\langle9,7,\text{-}7\rangle$
$\displaystyle \overrightarrow{AC} \:=\: \langle7, \text{-}1, \text{-}1\rangle$

Area of triangle $\displaystyle = \:\frac{\overrightarrow{AC} \times \overrightarrow{AB}}{2}$ . . . . Correct so far!

I end up getting vector $\displaystyle \langle \text{-}14, \text{-}40, \text{-}40\rangle$ . ← Here!

The correct answer is 35.9 unitsē.
Can someone help?

The cross product is: .$\displaystyle \langle\text{-}14, \text{-}40, {\color{blue}\text{-}58}\rangle$

Edit: Too fast for me, spec!
.