Why does $\vec{r}\,'(t)$ give the tangent vector to $\vec{r}(t)$ if $\vec{r}\,'(t)$ is orthogonal to $\vec{r}(t)$

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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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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).