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Prove some vector identities using the “cab minus bac” formula

In the previous exercise (Section 13.14, Exercise #9) we proved the “cab minus bac” formula:

    \[ A \times (B \times C) = (C \cdot A) B - (B \cdot A) C. \]

Using this formula prove the following identities:

  1. (A \times B) \times (C \times D) = (A \times B \cdot D)C - (A \times B \cdot C) D.
  2. A \times (B \times C) + B \times (C \times A) + C \times (A \times B) = O.
  3. A \times (B \times C) = (A \times B) \times C if and only if B \times (C \times A) = O.
  4. (A \times B) \cdot (C \times D) = (B \cdot D)(A \cdot C) - (B \cdot C)(A \cdot D).

  1. Proof. Using the cab minus bac formula with A \times B in place of A, C in place of B, and D in place of C we have

        \begin{align*}  (A \times B) \times (C \times D) &= (D \cdot (A \times B))C - (C \cdot (A \times B))D \\  &= (A \times B \cdot D) C - (A \times B \cdot C) D. \qquad \blacksquare \end{align*}

  2. Proof. Applying the cab minus back formula to each of the three terms in the sum we have

        \begin{align*}  A \times (B \times C) &= (C \cdot A) B - (B \cdot A)C \\  B \times (C \times A) &= (A \cdot B) C - (C \cdot B)A \\  C \times (A \times B) &= (B \cdot C) A - (A \cdot C)B. \end{align*}

    So, putting these together we have

        \begin{align*}  A &\times (B \times C) + B \times (C \times A) + C \times (A \times B) \\  &= (C \cdot A)B - (B\cdot A)C + (A \cdot B)C - (C \cdot B)A + (B \cdot C)A - (A \cdot C)B \\  &= O. \qquad \blacksquare \end{align*}

  3. Proof. From cab minus bac we have

        \[ A \times (B \times C) = (C \cdot A)B - (B \cdot A)C. \]

    Furthermore, since (A \times B) \times C = -C \times (A \times B), we can apply bac minus cab to get

        \begin{align*}  (A \times B) \times C &= -C \times (A \times B) &(\text{Thm 13.12(a)}) \\  &= - ((B \cdot C)A - (A \cdot C) B) \\  &= (A \cdot C)B - (B \cdot C)A \\  &= (C \cdot A)B - (B \cdot C)A + (A \cdot B)C - (A \cdot B)C \\  &= (C \cdot A)B - (B \cdot A)C + B \times (C \times A) \\  &= A \times (B \times C) + B \times (C \times A). \end{align*}

    Therefore,

        \[ (A \times B) \times C = A \times (B \times C) \iff B \times (C \times A) = O. \qquad \blacksquare\]

  4. Proof. From a previous exercise (Section 13.14, Exercise #7(d)) we know the identity A \cdot B \times C = C \cdot A \times B. In this case we have A \times B in place of A, C in place of B and D in place of C. This gives us

        \begin{align*}  (A \times B) \cdot (C \times D) &= D \cdot ((A \times B) \times C)\\  &= ((A \times B) \times C) \cdot D \\  &= (-C \times (A \times B)) \cdot D \\  &= (-(B \cdot C)A + (A \cdot C)B) \cdot D \\  &= (B \cdot D)(A \cdot C) - (B \cdot C)(A \cdot D). \qquad \blacksquare \end{align*}

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