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Prove some properties about a line and a point not on the line

by RoRi

Let Q be a point not on the line L(P;A) in \mathbb{R}^n.

  1. Consider the function

        \[ f(t) = \lVert Q - X(t) \rVert^2, \qquad \text{where} \qquad X(t) = P + tA. \]

    Prove f(t) is a quadratic polynomial in t and that there is a unique value of t, say t_0, at which this polynomial takes on its minimum.

  2. Prove that Q - X(t_0) is orthogonal to A.
  1. Proof. First, we compute

        \begin{align*} \lVert Q - X(t) \rVert^2 &= (Q - X(t)) \cdot (Q - X(t)) \\ &= (Q - P - tA) \cdot (Q - P - tA) \\ &= (Q-P)\cdot (Q-P) - t \big( (Q-P)\cdot A + A \cdot (Q-P) \big) + t^2 A \cdot A. \end{align*}

    But, (Q-P)\cdot (Q-P), \ (Q-P) \cdot A, \ A \cdot (Q-P), A \cdot A are all just real scalars (by definition of the dot product); therefore,

        \[ f(t) = c_1 - c_2 t + c_3 t^2 \]

    for scalars c_1, c_2, c_3 given by

        \[ c_1 = (Q-P) \cdot (Q-P), \quad c_2 = 2A \cdot (Q-P), \quad c_3 = A \cdot A. \]

    Then,

        \[ f'(t) = -c_2 + 2c_3 t \]

    which implies f(t) has a unique minimum at t_0 = \frac{c_2}{2c_3}. \qquad \blacksquare

  2. Proof. We compute the dot product,

        \begin{align*} (Q - X(t_0)) \cdot A &= (Q - P - t_0 A) \cdot A \\ &= (Q-P) \cdot A - t_0 A \cdot A. \end{align*}

    Then we have

        \[ t_0 = \frac{c_2}{2c_3} = \frac{2 A \cdot (Q-P)}{2A \cdot A}. \]

    Therefore,

        \[ (Q-X(t_0)) \cdot A = (Q-P) \cdot A - \frac{2A \cdot (Q-P)}{2A \cdot A} \cdot A \cdot A = 0. \qquad \blacksquare\]

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