Unit vector $n=a\hat{i}+b\hat{j}$ is perpendicular to the vector $(\hat{i}+\hat{j})$, then the value of a and b may be:

Since

$(a\hat{i}+b\hat{j})\perp (\hat{i}+\hat{j})$

Let $\hat{i}+\hat{j}=\vec{B}$

Also we know dot product of two vectors =0

Thus $(a\hat{i}+b\hat{j}).(\hat{i}+\hat{j})=0$

$a+b=0$

$a=-b$-----------(1)

As we know n=1 as n is a unit vector

$|n|=\sqrt{a^2+b^2}=1$

$a^2+b^2=1$

${(-b)}^2+b^2=1$

$b=\frac{1}{\sqrt{2}}$ and $a=\frac{-1}{\sqrt{2}}$