How do I compute the area of this parallelogram

Question

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"? — user247327, Jul 24 '16 at 12:34

score 2 · Accepted Answer · answered Jul 24 '16 at 20:33

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)

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

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

That is exactly how I solved it. THx – Ilan Aizelman WS Jul 27 '16 at 12:32 — Ilan Aizelman WS, Jul 27 '16 at 12:32
@IlanAizelmanWS excellent answer – user2323232 Aug 02 '16 at 16:22 — user2323232, Aug 02 '16 at 16:22

How do I compute the area of this parallelogram

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