Prove that F”(t) is orthogonal to F'(t) under given conditions

by RoRi

June 1, 2016

Let $B$ be a nonzero vector and $F$ a vector valued function with $F(t) \cdot B = t$ for all $t$ , and such that the angle between $F'(t)$ and $B$ is constant. Prove that $F''(t)$ and $F'(t)$ are orthogonal.

Proof. Since $F(t) \cdot B = t$ we have $F'(t) \cdot B = c$ for some constant $c$ . Since the angle between $F'(t)$ and $B$ is constant we have

$\frac{F'(t) \cdot B}{\lVert F'(t) \rVert \lVert B \rVert} = d$

for some constant $d$ . Therefore, $\lVert F'(t) \rVert$ is constant. Hence, $\lVert F'(t) \rVert^2$ is constant, say $a$ . So, we have

$\begin{align*} && \lVert F'(t) \rVert^2 &= a \\ \implies && F'(t) \cdot F'(t) &= a \\ \implies && ( F'(t) \cdot F'(t)) &= 0 \\ \implies && 2F'(t) \cdot F''(t) &= 0 \\ \implies && F'(t) \cdot F''(t) &= 0. \qquad \blacksquare \end{align*}$

2 comments

April 1, 2018 at 9:27 am

Van Gogh says:

Is F’(t)•B=1?

Reply
- February 4, 2020 at 4:01 pm
  
  Anonymous says:
  
  Yep, this is correct.
  
  Reply

Stumbling Robot

Prove that F”(t) is orthogonal to F'(t) under given conditions

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