Problem on dot products

$\newcommand{\bfA}{\mathbf{A}}$ $\newcommand{\bfB}{\mathbf{B}}$ $\newcommand{\bfC}{\mathbf{C}}$ $\newcommand{\bfF}{\mathbf{F}}$ $\newcommand{\bfI}{\mathbf{I}}$ $\newcommand{\bfa}{\mathbf{a}}$ $\newcommand{\bfb}{\mathbf{b}}$ $\newcommand{\bfc}{\mathbf{c}}$ $\newcommand{\bfd}{\mathbf{d}}$ $\newcommand{\bfe}{\mathbf{e}}$ $\newcommand{\bfi}{\mathbf{i}}$ $\newcommand{\bfj}{\mathbf{j}}$ $\newcommand{\bfk}{\mathbf{k}}$ $\newcommand{\bfn}{\mathbf{n}}$ $\newcommand{\bfr}{\mathbf{r}}$ $\newcommand{\bfu}{\mathbf{u}}$ $\newcommand{\bfv}{\mathbf{v}}$ $\newcommand{\bfw}{\mathbf{w}}$ $\newcommand{\bfx}{\mathbf{x}}$ $\newcommand{\bfy}{\mathbf{y}}$ $\newcommand{\bfz}{\mathbf{z}}$
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
    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 $\mathbf{0} \neq \bfj.$