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If $\mathbf{a} \perp \mathbf{b}$ and $(\mathbf{a}+\mathbf{b}) \perp(\mathbf{a}+m \mathbf{b})$, then $m$ is equal to
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Verified Answer
The correct answer is:
$\frac{-|a|^{2}}{|b|^{2}}$
If $\mathbf{a}$ and $\mathbf{b}$ are perpendicular to each other.
$\begin{array}{ll}
\text { Then, } & a \cdot b=0 \\
\because & (a+b) \perp(a+m b) \\
\therefore & (a+b) \cdot(a+m b)=0 \\
\Rightarrow & a \cdot a+m b \cdot b+b \cdot a+m a \cdot b=0 \\
\Rightarrow \quad & |a|^{2}+m|b|^{2}+0+m \cdot 0=0 \\
\Rightarrow \quad & m=-\frac{|a|^{2}}{|b|^{2}}
\end{array}$
$\begin{array}{ll}
\text { Then, } & a \cdot b=0 \\
\because & (a+b) \perp(a+m b) \\
\therefore & (a+b) \cdot(a+m b)=0 \\
\Rightarrow & a \cdot a+m b \cdot b+b \cdot a+m a \cdot b=0 \\
\Rightarrow \quad & |a|^{2}+m|b|^{2}+0+m \cdot 0=0 \\
\Rightarrow \quad & m=-\frac{|a|^{2}}{|b|^{2}}
\end{array}$
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