A twist on finding angle between planes

Solution .

$\vec{r}(t) = \left \langle 2t, t^2, \dfrac{t^3}{3} \right \rangle$ $\vec{T}(t) = \dfrac{\vec{r}’(t)}{\left|\vec{r}’(t) \right|} = \dfrac{\left \langle 2, 2t, t^2 \right \rangle}{t^2 +2}$ $\vec{N}(t) = \dfrac{\vec{T}’(t)}{\left|\vec{T}’(t) \right|}$ $\vec{T}’(t) = \left \langle -2(t^2 +2)^{-2} 2t, \dfrac{2(t^2 +2)-(2t)^2}{(t^2 +2)^2}, \dfrac{2t(t^2 + 2)- 2t^3}{(t^2 +2)^2}, \right \rangle = \left \langle \dfrac{-4t}{(t^2 +2)^2}, \dfrac{4-2t^2}{(t^2 +2)^2}, \dfrac{4t}{(t^2 +2)^2} \right \rangle$ $\left| \vec{T}’(t) \right| = \sqrt{\dfrac{16t^2 + 16 + 4t^4 - 16t^2 + 16t^2}{(t^2 + 2)^4}} = \dfrac{2\sqrt{t^4 + 4t^2 + 4}}{(t^2 +2)^2} = \dfrac{2}{t^2 +2}$ $\vec{N}(t) = \dfrac{\left \langle -2t, 2-t^2, 2t\right \rangle}{t^2 +2}$

The binormal vector $\vec{B}(t)$ will be a normal vector to the osculating plane at time $t$ given by vectors $\vec{N}(t), \vec{T}(t)$ since $\vec{B}(t) = \vec{N}(t) \times \vec{T}(t)$

$\vec{B}(t) = \vec{N}(t) \times \vec{T}(t) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \dfrac{-2t}{t^2 + 2} &\dfrac{2-t^2}{t^2 + 2} &\dfrac{2t}{t^2 + 2} \\ \dfrac{2}{t^2 + 2}&\dfrac{2t}{t^2 + 2} & \dfrac{t^2}{t^2 + 2} \end{vmatrix}$

$\vec{B}(t) = \hat{i} \left[ \left(\dfrac{2-t^2}{t^2 +2} \right) \left( \dfrac{t^2}{t^2 +2}\right) - \left(\dfrac{2t}{t^2 +2} \right) \left(\dfrac{2t}{t^2 +2} \right) \right] - \hat{j}\left[ \left(\dfrac{-2t}{t^2 +2} \right) \left( \dfrac{t^2}{t^2 +2}\right) - \left(\dfrac{2t}{t^2 +2} \right) \left( \dfrac{2}{t^2 +2}\right) \right] + \hat{k}\left[ \left(\dfrac{-2t}{t^2 +2} \right) \left( \dfrac{2t}{t^2 +2}\right) - \left(\dfrac{2-t^2}{t^2 +2} \right) \left( \dfrac{2}{t^2 +2}\right) \right]$

$= \dfrac{t^2 \hat{i} + 2t \hat{j} - 2\hat{k}}{t^2 +2}$ $\vec{B}(1) = \left \langle \dfrac{-1}{3}, \dfrac{2}{3}, \dfrac{-2}{3} \right \rangle; \vec{B}(2) = \left \langle \dfrac{-2}{3}, \dfrac{2}{3}, \dfrac{-1}{3} \right \rangle$

The angle between two planes is the same as the angle between their normal vectors.

$\cos \theta = \dfrac{\vec{B}(1) \cdot \vec{B}(2)}{\left|\vec{B}(1) \right| \left| \vec{B}(2) \right| } = \dfrac{\dfrac{-1}{3} \cdot \dfrac{-2}{3} + \dfrac{2}{3} \cdot \dfrac{2}{3} + \dfrac{-2}{3} \cdot \dfrac{-1}{3} }{\sqrt{ \left( \dfrac{-1}{3} \right)^2 + \left( \dfrac{2}{3} \right)^2 + \left( \dfrac{-2}{3} \right)^2} \sqrt{ \left( \dfrac{-2}{3} \right)^2 + \left( \dfrac{2}{3} \right)^2 + \left( \dfrac{-1}{3} \right)^2} } = \dfrac{8}{9}; \arccos \left( \dfrac{8}{9} \right) = \theta$

$\alpha + \beta = 8+9=\boxed{17}$