0
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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"? $\endgroup$
    – user247327
    Jul 24, 2016 at 12:34
  • $\begingroup$ sides, find the area. @user247327 $\endgroup$ Jul 24, 2016 at 12:36
  • $\begingroup$ Okay. solved it. it equals to 21 $\endgroup$ Jul 24, 2016 at 12:47

1 Answer 1

2
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ That is exactly how I solved it. THx $\endgroup$ Jul 27, 2016 at 12:32
  • 1
    $\begingroup$ @IlanAizelmanWS excellent answer $\endgroup$ Aug 2, 2016 at 16:22

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .