0
$\begingroup$

In 1-4 of Do Carmo's Curves and Surfaces, he states that so long as $\mathbf{u} \land \mathbf{v} \neq 0$ for two vectors $\mathbf{u}$ and $\mathbf{v}$ (where $\land$ denotes the cross product between vectors:

First, we observe that $(\mathbf{u}\land \mathbf{v})·(\mathbf{u}\land \mathbf{v}) = |\mathbf{u}\land \mathbf{v}|^2 > 0$. This means that the determinant of the vectors $\mathbf{u},\mathbf{v},\mathbf{u \land v}$ is positive; that is, $\{\mathbf{u},\mathbf{v},\mathbf{u \land v}\}$ is a positive basis.

Why is this true? How do we know from the "First..." that the determinant is positive?

$\endgroup$
5
  • 1
    $\begingroup$ I don't have Do Carmo's book but often the determinant WRT an oriented orthonormal basis $B$ is used to define the vector product via $\langle u \wedge v \mid \bullet \rangle =\det_B(u,v,\bullet)$. Which explains the statement if you replace $\bullet$ with $u \wedge v$. $\endgroup$
    – Olivier Bégassat
    Commented Jun 15 at 13:36
  • $\begingroup$ @OlivierBégassat could you provide some more detail as to how that explains the statement? And what does the $\langle | \rangle$ notation mean? $\endgroup$
    – DC2974
    Commented Jun 15 at 13:41
  • $\begingroup$ It's the scalar product i.e. dot product. $\endgroup$
    – Olivier Bégassat
    Commented Jun 15 at 13:41
  • $\begingroup$ Have you tried the substitution? $\endgroup$
    – Olivier Bégassat
    Commented Jun 15 at 13:41
  • 1
    $\begingroup$ Ah yes, that makes total sense. Thank you! $\endgroup$
    – DC2974
    Commented Jun 15 at 13:43

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .