KCET · Maths · Vector Algebra
If \(\mathbf{a} \perp \mathbf{b}\) and \((\mathbf{a}+\mathbf{b}) \perp(\mathbf{a}+m \mathbf{b})\), then \(m\) is equal to
- A \(-1\)
- B 1
- C \(\frac{-|a|^{2}}{|b|^{2}}\)
- D 0
Answer & Solution
Correct Answer
(C) \(\frac{-|a|^{2}}{|b|^{2}}\)
Step-by-step Solution
Detailed explanation
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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