Answer:
Option D
Explanation:
We have |a|=3,|b|=4 and |a+b|=5
Since, |a+b|=5
$\therefore$ $|a+b|^{2}=25$
$\Rightarrow$ $(a+b).(a+b)=25$
$\Rightarrow$ $a.a+a.b+b+a+b.b=25$
$\Rightarrow$ $|a|^{2}+2a.b+|b|^{2}=25$ [$\therefore$ a.b=b.a]
$\Rightarrow$ $9+2 a.b+16=25$
$\Rightarrow$ $a.b=0$
Now, consider $|a-b|^{2}=(a-b)(a-b)$
= $a.a-a.b-b.a+b.b$
= $|a|^{2}-2 a.b+|b|^{2}$ [$\because$ a.b=b.a]
=$9-0+16=25$
$\Rightarrow$ $|a-b|=5$
