If a → and b → are two vectors such that | a → | = 2 , | b → | = 3 and a → + t…

We know that, Dot product of perpendicular vectors is zero.

$\therefore \left(\overrightarrow{a}+t\overrightarrow{b}\right)·\left(\overrightarrow{a}-t\overrightarrow{b}\right)=0$

$\Rightarrow {\left|\overrightarrow{a}\right|}^2-t\left(\overrightarrow{a}·\overrightarrow{b}\right)+t\left(\overrightarrow{b}·a\right)-t^2{\left|\overrightarrow{b}\right|}^2=0$

$\Rightarrow {\left|\overrightarrow{a}\right|}^2-t^2{\left|\overrightarrow{b}\right|}^2=0$

$\Rightarrow 4-9t^2=0$

$\Rightarrow t^2=\frac{4}{9}$

$\Rightarrow t=\pm \frac{2}{3}$

But given that $t$ is positive scalar, hence $t=\frac{2}{3}$.