The unit vector (a hat{imath} + b hat{jmath}) | vedclass

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(D) Let the unit vector be $\hat{n} = a \hat{\imath} + b \hat{\jmath}$ and the given vector be $\vec{r} = \hat{\imath} + \hat{\jmath}$.Since $\hat{n}$

Vedclass · Accepted Answer

$-\frac{1}{\sqrt{2}}$

Vedclass · Answer

(D) Let the unit vector be $\hat{n} = a \hat{\imath} + b \hat{\jmath}$ and the given vector be $\vec{r} = \hat{\imath} + \hat{\jmath}$.Since $\hat{n}$ is perpendicular to $\vec{r}$,their dot product must be zero: $\hat{n} \cdot \vec{r} = 0$.$(a \hat{\imath} + b \hat{\jmath}) \cdot (\hat{\imath} + \hat{\jmath}) = a + b = 0$,which implies $b = -a$.Since $\hat{n}$ is a unit vector,its magnitude is $1$: $\sqrt{a^2 + b^2} = 1$.Substituting $b = -a$,we get $\sqrt{a^2 + (-a)^2} = 1$,which simplifies to $\sqrt{2a^2} = 1$,or $|a|\sqrt{2} = 1$.Thus,$a = \pm \frac{1}{\sqrt{2}}$.If $a = \frac{1}{\sqrt{2}}$,then $b = -\frac{1}{\sqrt{2}}$.If $a = -\frac{1}{\sqrt{2}}$,then $b = \frac{1}{\sqrt{2}}$.Comparing with the given options,the value $b = -\frac{1}{\sqrt{2}}$ is present as option $D$.

The unit vector $(a \hat{\imath} + b \hat{\jmath})$ is perpendicular to $(\hat{\imath} + \hat{\jmath})$. The value of $b$ is

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