Question

If $a\perp b$ and $(a+b)\perp (a+mb)$, then m is equal to

• A. $-1$
• B. $1$
• C. $\frac{-{|a|}^2}{{|b|}^2}$
• D. $0$

Answer 1

Answer: (C)

If $a$ and $b$ are perpendicular to each other.
Then, $a\cdot b=0...(i)$
$\because (a+b)\perp (a+mb)$
$\therefore (a+b)\cdot (a+mb)=0$
$\Rightarrow a\cdot a+mb\cdot b+b\cdot a+ma\cdot b=0$
$\Rightarrow |a{|}^2+m|b{|}^2+0+m\cdot 0=0$[from Eq. (i)]
$\Rightarrow m=-\frac{|a{|}^2}{|b{|}^2}$