# Components of the vector product

consider a parallelogram $ABCD$ with $A(3,-2,-1)$, $B(2,1,3)$ and $C(0,4,1)$ find the coordinates of $D$ and calculate the area of this parallelogram

a parallelogram can be divided into two triangles of equal area by a diagonal,that is if you find area of triangle ABC and then multiply it by two, you will get area of parallelogram. area of triangle is half of product of two sides and angle between them, in your case area of triangle is 1/2{(AB)(BC)sin*} where * is the angle between A and B, and you know that (AB)(BC)sin* = ABxBC(cross product), twice of area of this triangle is area of your parallelogram,hence area of parallelogram becomes(ABxBC).solve ABxBC and get the answer.

• guys i don't know how to use mathjex, so if anyone is having a problem in understanding my answer please leave a comment. Sep 10, 2016 at 9:17

$$\vec{OD}=\vec{OA}+\vec{AB}+\vec{AC}.$$

For the area, you know (I hope) that the altitude is given by $$Alt=\left\|\vec{AB}-\frac{\left<\vec{AB},\vec{AC}\right>}{\|\vec{AC}\|^2}\right\|.$$ I let you conclude.

$$\quad{\text{Area=}}\overrightarrow{AB}\times \overrightarrow{BC}$$ $$\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{DC}+\overrightarrow{AD}=0$$

Find the vectors C-B = (0-2 ; 4-1 ; 1-3) =(-2 ; 3 ; -2) and A-B = (3-2 ; -2 -1 ; -1 -3)= (1 ; -3 ; -4) Now you can do the cross product between C-B and A-B and the results is the area of the parallelogram (because the parrallelogram is two triangles. The area of a triangle is 0.5 for the product of two sides for the sin of the angle between those two sides that is the cross product). I remember you that you must find the magnitude of the vector getted from the cross product for find the area. In this way you do not need to calculate the coordinate D