The unit vector $(a \hat{\imath}+b \hat{j})$ is perpendicular to $(\hat{\imath}+\hat{\jmath})$. The value of…
The unit vector $(a \hat{\imath}+b \hat{j})$ is perpendicular to $(\hat{\imath}+\hat{\jmath})$. The value of $^{\prime} b^{\prime}$ is
$+\frac{1}{\sqrt{3}}$
$-\frac{1}{\sqrt{3}}$
$+\frac{1}{2}$
$-\frac{1}{\sqrt{2}}$
Solution
$\begin{array}{l}
\hat{\mathrm{n}}=a \hat{\mathrm{i}}+b \hat{\mathrm{j}} \\
\overrightarrow{\mathrm{r}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}
\end{array}$
If $\hat{n}$ is perpendicular to the vector $\vec{r}=\hat{i}+\hat{j}$
$\hat{n} \cdot \vec{r}=0$
$a+b=0$
$b=-a . \ldots . .(1)$
The magnitude of unity vector is $1 .$
$\sqrt{a^{2}+b^{2}}=1$ $\sqrt{a^{2}+a^{2}}=\sqrt{2} a=1($ from equation 1) $a=\frac{1}{\sqrt{2}}$
$a=\frac{1}{\sqrt{2}}$
and from equation (1), we get
$b=-\frac{1}{\sqrt{2}}$
The correct option is D.