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Prove that F”(t) is orthogonal to F'(t) under given conditions

Let B be a nonzero vector and F a vector valued function with F(t) \cdot B = t for all t, and such that the angle between F'(t) and B is constant. Prove that F''(t) and F'(t) are orthogonal.


Proof. Since F(t) \cdot B = t we have F'(t) \cdot B = c for some constant c. Since the angle between F'(t) and B is constant we have

    \[ \frac{F'(t) \cdot B}{\lVert F'(t) \rVert \lVert B \rVert} = d \]

for some constant d. Therefore, \lVert F'(t) \rVert is constant. Hence, \lVert F'(t) \rVert^2 is constant, say a. So, we have

    \begin{align*}  && \lVert F'(t) \rVert^2 &= a \\  \implies && F'(t) \cdot F'(t) &= a \\  \implies && ( F'(t) \cdot F'(t)) &= 0 \\  \implies && 2F'(t) \cdot F''(t) &= 0 \\  \implies && F'(t) \cdot F''(t) &= 0. \qquad \blacksquare \end{align*}

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