# How do I compute the area of this parallelogram

Given vectors $a,b$ and the ribs of parallelogram are $2a +3b = A$, $a-2b = B$.

Also given $a \times b = (-1,2,2)$. Compute the surface of the parallelogram.

I'm not sure where I saw but I think it should be something along the line $\frac{1}{6} A \times B = S_{parallelogram}$. and if thats true. I'm not sure how to compute that determinant.

Any hints are highly appreciated.

• What do you mean by the "ribs" of a parallelogram? The sides or the diagonals? And by "compute the surface" do you mean "find the area"? Jul 24, 2016 at 12:34
• sides, find the area. @user247327 Jul 24, 2016 at 12:36
• Okay. solved it. it equals to 21 Jul 24, 2016 at 12:47

For two vectors $v,w \in \mathbb{R}^3$, the area of the parallelogram is the square root of the Gram determinant $\text{Gram}(v,w)$. For two three-dimensional vectors you conveniently have

$$\sqrt{\text{Gram}(v,w)} = ||v \times w||$$

where $||.||$ is the euclidian norm.

So in your case you get

$$Vol_2(P(2a+3b, a-2b)) = ||(2a+3b) \times (a-2b)||$$

with (cross product is bilinear, alternating and anti-commutative)

$$$$(2a+3b) \times (a-2b) = ((2a+3b) \times a) - ((2a+3b) \times 2b) = \\ = \underbrace{(2a \times a)}_{=0} + \underbrace{(3b \times a)}_{=-3 a \times b} - \underbrace{(2a \times 2b)}_{4 a \times b} - \underbrace{(3b \times 2b)}_{=0} = \\ = -7 a \times b = (7, -14, -14)$$$$

thus giving the result $$||(7,-14,-14)|| = \sqrt{49+196+196} = 21$$

• That is exactly how I solved it. THx Jul 27, 2016 at 12:32
• @IlanAizelmanWS excellent answer Aug 2, 2016 at 16:22