Area of a parallelogram in $ \mathbb{R}3 $ Let $ A = (2, -3, 1), B = (5, -3, -1), C = (-2, -3, 5) $ and $ D = (1, -3, 3) $
I have the above situation. I know this: $$ A = \| \vec{AB} \times \vec{AD} \| $$ gives the area of the parallelogram.
My end components are: $ 0\hat{i} - 4\hat{j} + 0\hat{k} = -4\hat{j} $
The area I get is $4$ for this... but it's wrong and don't know where I'm going wrong.

The answer is indeed 4. The answer that was given was wrong.
