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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Similar Questions

Can you associate vectors with $(a)$ the length of a wire bent into a loop,$(b)$ a plane area,$(c)$ a sphere? Explain.

Which of the following colours of visible light has the lowest energy?

Let $\theta$ be the angle between vectors $\vec{A}$ and $\vec{B}$. Which of the following figures correctly represents the angle $\theta$?

Two forces $F_1 = 1\,N$ and $F_2 = 2\,N$ act along the lines $x = 0$ and $y = 0$ respectively. Then the resultant of forces would be

Which of the following statement$(s)$ is/are correct?

Vedclass Test Series

Exam Paper Generator

Online Exam Module