I have seen a proof that $\vec{r}\,'(t)$ is orthogonal to $\vec{r}(t)$, but $\vec{r}\,'(t)$ gives the tangent vector to the curve $\vec{r}(t)$, for any $t$. I don't understand how $\vec{r}\,'(t)$ can represent the tangent vector but also be orthogonal to the vector curve?

Just another question related to the above: If the binormal vector is defined to be $$\vec{B}(t) = \vec{T}(t) \times \vec{N}(t),$$ where $\vec{T}(t)$ and $\vec{N}(t)$ represent the unit tangent and unit normal vector respectively, does the binormal vector give a unit vector orthogonal to both the tangent and normal vector, in the direction according to the 'right hand rule'? What significance does this have?

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    $\begingroup$ In general the derivative of the position vector is not orthogonal to the position vector. Are you thinking of the derivative of the unit vector being orthogonal to the position vector? $\endgroup$
    – Andrew
    Commented Mar 7, 2019 at 22:15
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    $\begingroup$ Consider a circle: in such a case, the tangent vector is orthogonal to the position vector. $\endgroup$ Commented Mar 7, 2019 at 22:19
  • $\begingroup$ @Andrew I was just on this page, where it gives a fact about the orthogonality about the derivative of a position vector, but I don't think I understand it. $\endgroup$
    – Gurjinder
    Commented Mar 7, 2019 at 22:25
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    $\begingroup$ That is assuming that $\left\|\vec{r}(t)\right\|$ is constant over time. In general though, $\vec{r}(t)$ need not be orthogonal to $\vec{r}'(t)$. However, it is certainly possible to happen, for example if $\vec{r}(t)$ is tracing out a circle centred at the origin. $\endgroup$ Commented Mar 7, 2019 at 22:27
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    $\begingroup$ @MinusOne-Twelfth , like a circle, centred at the origin? $\endgroup$
    – Gurjinder
    Commented Mar 7, 2019 at 22:27

1 Answer 1


To help make sense of this think of $r(t)$ as your position as you are driving around on earth (approximated as a sphere). The tangent vector $T(t)$ is the direction your car is facing. The normal vector $N(t)$ is the direction you're turning your steering wheel (left or right). The binormal vector is a vector orthogonal to both (pointing up or down).


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