Problem on dot products

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If $\bfx \cdot \bfy = \bfx \cdot \bfz$ for a nonzero $\bfx$, does $\bfy = \bfz$? If so, prove it. If not, provide a counterexample.

  • Solution

    For problems where we determine if a statement is true, we should start by exploring the problem in simple cases.
    Let's choose simple values for $\bfx, \bfy, \bfz.$ Because $\bfx$ must be nonzero, a simple choice is $\bfx = \bfi$. There is no constraint on $\bfy$, so take $\bfy = \mathbf{0}.$
    Noting that $\bfx \cdot \bfy=0$ in this case, the problem becomes:
    If $0 = \bfi \cdot \bfz$, does $\bfz = \mathbf{0}$?
    Recall that
    The dot product of perpendicular vectors is zero.
    Hence, any vector $\bfz$ that is perpendicular to $\bfi$ will provide a counterexample, such as $\bfz = \bfj$.
    The statement is false because $\bfi \cdot \mathbf{0} = \bfi \cdot \bfj$, but $\mathbf{0} \neq \bfj.$

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