Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

There is a regular parametric curve $r(t)$ in $\Bbb{R}^3$,and $r'(t)\bullet a=0$,where $a$ is a fixed vector in $\Bbb{R}^3$, show that $r(t)$ is a plane curve. Thanks very much

share|cite|improve this question
    
You assumed $r(t)=x(t)i+y(t)j+z(t)k$ an smooth function? – S. Snape Aug 28 '13 at 8:15
    
@BabakS. Yes, $r(t)$ is at least $\in C^3(\Bbb{R}^3)$ – Laura Aug 28 '13 at 8:21
up vote 2 down vote accepted

Let $(a,b,c)$ be an orthonormal basis of $\mathbb{R}^3$ and write $r(t)=x(t)a+y(t)b+z(t)c$. Because $r'(t) \cdot a=0$, we deduce $x'(t)=0$ that is $x(t)$ is constant. Therefore, the curve lies on an affine plane parallel to the plane generated by $(b,c)$.

share|cite|improve this answer

Another way to view it: let $L(u) = a\bullet u$, and put $f(t) = L(r(t))$. Then $f'(t) = a \bullet r'(t) = 0$, which shows that $L$ is constant along the curve. In other words, the curve is contained in a level set of $L$, i.e. in a plane with normal vector $a$.

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.