Problem:
Solution:
The tangent line has direction $ \vec{a} = (3, 4t, 8t^2) $
The given line has parametric form $ (u, 0, u) $ so it's direction $ \vec{b} (1, 0, 1) $
The angle between these two direction is $ \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \frac{(3 + 8t^2)}{\sqrt{3^2 + (4t)^2 + (8t^2)^2}\sqrt{1^2 + 0^2 + 1^2}} = \frac{(3 + 8t^2)}{\sqrt{9 + 16t^2 + 64t^4}\sqrt{2}} = \frac{(3 + 8t^2)}{\sqrt{(3 + 8t^2)^2}\sqrt{2}} = \frac{1}{\sqrt{2}}$
So the angle is constant.
Solution:
The tangent line has direction $ \vec{a} = (3, 4t, 8t^2) $
The given line has parametric form $ (u, 0, u) $ so it's direction $ \vec{b} (1, 0, 1) $
The angle between these two direction is $ \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \frac{(3 + 8t^2)}{\sqrt{3^2 + (4t)^2 + (8t^2)^2}\sqrt{1^2 + 0^2 + 1^2}} = \frac{(3 + 8t^2)}{\sqrt{9 + 16t^2 + 64t^4}\sqrt{2}} = \frac{(3 + 8t^2)}{\sqrt{(3 + 8t^2)^2}\sqrt{2}} = \frac{1}{\sqrt{2}}$
So the angle is constant.
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