The expression left( frac{1}{sqrt{2}}hat{i} + | vedclass

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

Question

(A) Let the given vector be $\vec{P} = \frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j}$.The magnitude of a vector $\vec{A} = x\hat{i} + y\hat{j}

Vedclass · Accepted Answer

Unit vector

Vedclass · Answer

(A) Let the given vector be $\vec{P} = \frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j}$.The magnitude of a vector $\vec{A} = x\hat{i} + y\hat{j}$ is given by $|\vec{A}| = \sqrt{x^2 + y^2}$.Substituting the values,we get $|\vec{P}| = \sqrt{\left( \frac{1}{\sqrt{2}} \right)^2 + \left( \frac{1}{\sqrt{2}} \right)^2}$.$|\vec{P}| = \sqrt{\frac{1}{2} + \frac{1}{2}} = \sqrt{1} = 1$.Since the magnitude of the vector is $1$,it is a unit vector.

The expression $\left( \frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j} \right)$ is a

Similar Questions

$\vec{A}$ is a vector of magnitude $2.7$ units due east. What is the magnitude and direction of vector $4 \vec{A}$?

What is the energy of an electron in a hydrogen atom in a stationary state corresponding to $n = 2$ ?

$ABC$ is an equilateral triangle. The length of each side is $a$ and the centroid is point $O$. Find $\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CA} = \ldots \ldots$.

If $|\vec{P}+\vec{Q}|=|\vec{P}|=|\vec{Q}|$,then the angle between $\vec{P}$ and $\vec{Q}$ is: (in $^{\circ}$)

The angle between vectors $\vec a$ and $\vec b$ is $\frac{\pi }{6}$. The angle between vectors $-3\vec a$ and $2\vec b$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module