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Prove that if A x B = O and A . B = 0 then A or B is O.

  1. Prove that if A \times B = O and A \cdot B = 0, then at least one of A,B is the zero vector. Give a geometric interpretation of this result.
  2. Prove that if A is a nonzero vector, and if A \times B = A \times C and A \cdot B = A \cdot C then B = C.

  1. Proof. If we have A \times B = O then

        \begin{align*}  A \times B = O && \implies && \lVert A \times B \rVert &= 0 \\  && \implies && \lVert A \rVert \lVert B \rVert \sin \theta &= 0. \end{align*}

    But, since A \cdot B = 0 we know A and B are orthogonal, so \sin \theta = 1. Therefore, we must have either \lVert A \rVert = 0 or \lVert B \rVert = 0. Hence, either A or B is the zero vector. \qquad \blacksquare

  2. Proof. First,

        \[ A \times B = A \times C \quad \implies \quad A \times B - A \times C = O \quad \implies \quad A \times (B -C) = O. \]

    Then,

        \[ A \cdot B = A \cdot C \quad \implies \quad A \cdot B - A \cdot C = 0 \quad \implies \quad A \cdot (B-C) = 0. \]

    But, A \times (B-C) = O and A \cdot (B-C) = 0 implies A = O or B-C = O (by part (a)). Since A \neq O by hypothesis, we must have B-C = O. Hence, B = C. \qquad \blacksquare

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