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Prove statements about a vector satisfying a given vector equation

Consider unit length, orthogonal vectors A,B \in \mathbb{R}^3, and a vector P such that

    \[ P \times B = A-P. \]

Prove the following.

  1. P and B are orthogonal and the length of P is \frac{1}{2} \sqrt{2}.
  2. The vectors P,B,P \times B form a basis for \mathbb{R}^3.
  3. (P \times B) \times B = -P.
  4. P = \frac{1}{2}A - \frac{1}{2} (A \times B).

  1. Proof. We compute,

        \begin{align*}  P \times B = A - P && \implies && B \cdot (P \times B) &= B \cdot A - B \cdot P \\  && \implies && 0 = B \cdot P \end{align*}

    since B \cdot (P \times B) = 0 and B \cdot A = 0 since A and B are orthogonal by assumption. Thus, B and P are orthogonal. Next,

        \begin{align*} \lVert P \times B \rVert^2 &= \lVert P\rVert^2 \lVert B \rVert^2 - (P \cdot B)^2 \\  &= \lVert P \rVert^2 \end{align*}

    since \lVert B \rVert^2 = 1 by hypothesis and P \cdot B = 0. Hence, from the vector equation we have

        \begin{align*}  P \times B = A - P && \implies && \lVert P \times \rVert^2 &= \lVert A-P \rVert^2 \\  && \implies && \lVert P \rVert^2 &= \lVert A \rVert^2 - \lVert P \rVert^2 \\  && \implies && 2 \lVert P \rVert^2 &= \lVert A \rVert^2 \\  && \implies && 2 \lVert P \rVert^2 &= 1 \\  && \implies && \lVert P \rVert^2 &= \frac{1}{2} \\  && \implies && \lVert P \rVert &= \frac{\sqrt{2}}{2}. \qquad \blacksquare \end{align*}

  2. Proof. Since P and B are orthogonal (part (a)), we know the vectors P,B, P \times B are independent. Thus, they form a basis for \mathbb{R}^3 since any three independent vectors in \mathbb{R}^3 are a basis. \qquad \blacksquare
  3. Proof. We compute, the vector (P \times B) \times B is given by

        \[ (p_2 b_3 - p_3 b_2, p_3 b_1 - p_1 b_3, p_1 b_2 - p_2 b_1) \times B. \]

    Then the three coordinates of this cross product are given by

        \begin{align*}  &(p_3 b_1 - p_1 b_3)b_3 - (p_1 b_2 - p_2 b_1)b_2 \\  &(p_1 b_2 - p_2 b_1)b_1 - (p_2 b_3 - p_3 b_2)b_3 \\  &(p_2 b_3 - p_3 b_2)b_2 - (p_3 b_1 - p_1 b_3)b_1. \end{align*}

    Expanding these out we obtain the coordinates

        \begin{align*}  &p_3 b_1 b_3 - p_1 b_3^2 - p_1 b_2^2 + p_2 b_1 b_2 \\  &p_1 b_1 b_2 - p_2 b_1^2 - p_2 b_3^2 + p_3 b_2 b_3 \\  &p_2 b_2 b_3 - p_3 b_2^2 - p_3 b_1^2 + p_1 b_1 b_3. \end{align*}

    Since \lVert B \rVert = 1 we know b_1^2 + b_2^2 + b_3^2 = 1 and since P \cdot B = 0 we know p_1 b_1 + p_2 b_2 + p_3 b_3 = 0. So, simplifying the expressions above, for each of the coordinates we have

        \begin{align*}  p_3 b_1 b_3 - p_1 b_3^2 - p_1 b_2^2 + p_2 b_1 b_2 &= p_3 b_1 b_3 - p_1 (1-b_1^2) + p_2 b_1 b_2 \\   &= p_3 b_1 b_3 - p_1 + p_1 b_1^2 + p_2 b_1 b_2 \\  &= b_1(p_3 b_3 + p_1 b_1 + p_2 b_2) - p_1 \\  &= -p_1 \\[9pt]  p_1 b_1 b_2 - p_2 b_1^2 - p_2 b_3^2 + p_3 b_2 b_3 &= p_1 b_1 b_2 - p_2 (1-b_2^2) + p_3 b_2 b_3 \\  &= p_1 b_1 b_2 - p_2 + p_2 b_2^2 + p_3 b_2 b_3 \\  &= b_2(p_1 b_1 + p_2 b_2 + p_3 b_3) - p_2 \\  &= -p_2 \\[9pt]  p_2 b_2 b_3 - p_3 b_2^2 - p_3 b_1^2 + p_1 b_1 b_3 &= p_2 b_2 b_3 - p_3 (1-b_3^2) + p_1 b_1 b_3 \\  &= p_2 b_2 b_3 - p_3 + p_3 b_3^2 + p_1 b_1 b_3 \\  &= b_3 (p_2 b_2 + p_3 b_3 + p_1 b_3) - p_3 \\  &= -p_3. \end{align*}

    Hence, we indeed have

        \[ (P \times B) \times B = (-p_1, -p_2, -p_3)  = -P. \qquad \blacksquare \]

  4. Proof. We compute

        \begin{align*}  (P \times B) \times B = -P && \implies && (A-P) \times B &= -P \\  && \implies && (A \times B) - (P \times B) &= -P \\  && \implies && (A \times B) - A + P &= -P \\  && \implies && -2P &= (A \times B) - A \\  && \implies && P &= \frac{1}{2}A - \frac{1}{2}(A \times B). \qquad \blacksquare \end{align*}

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