Why this equation is correct? (Frenet frame)

Question

$2(T_a-T_b)•(T'_a-T'_b)=-2T_a•T'_b-2T'_a•T_b$

Is $2T_a•T'_a=0$ ?

Please explain me why this equation work.

Answer 1

The equation

$(T_a - T_b) \cdot (T_a' - T_b')$ $= T_a \cdot (T_a' - T_b') - T_b \cdot (T_a' - T_b') = T_a \cdot T_a' - T_a \cdot T_b' - T_b \cdot T_a' + T_b \cdot T_b' \tag 1$

follows immediately from the ordinary distributive properties of the dot product $A \cdot B$ for any vectors $A$ and $B$.

If we understand $T_s$ to be the unit tangent vector field along some arc-length parametrized regular curve $\gamma(s)$ in the sense of Frenet-Serret,

$T_s = \gamma'(s), \tag 2$

then since

$T_s \cdot T_s = 1, \tag 3$

we find

$2T_s \cdot T_s' = T_s' \cdot T_s + T_s \cdot T_s' = 0 \tag 4$

holds for any value of $s$; taking $s = a$ and $s = b$, we have

$T_a \cdot T_a' = T_b \cdot T_b' = 0; \tag 5$

thus (1) becomes

$(T_a - T_b) \cdot (T_a' - T_b') = - T_a \cdot T_b' - T_b \cdot T_a', \tag 6$

which we may multiply by $2$ to obtain

$2(T_a - T_b) \cdot (T_a' - T_b') = 2(- T_a \cdot T_b' - T_b \cdot T_a') = -2T_a \cdot T_b' - 2T_b \cdot T_a'. \tag 7$