If vectors overrightarrow {A} = cosomega that | vedclass

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

Question

(D) Two vectors $\overrightarrow {A}$ and $\overrightarrow {B}$ are orthogonal to each other if their scalar product is zero,i.e.,$\overrightarrow {A}

Vedclass · Accepted Answer

$t=\frac{\pi}{\omega}$

Vedclass · Answer

(D) Two vectors $\overrightarrow {A}$ and $\overrightarrow {B}$ are orthogonal to each other if their scalar product is zero,i.e.,$\overrightarrow {A} \cdot \overrightarrow {B} = 0$.Given $\overrightarrow {A} = \cos \omega t \hat i + \sin \omega t \hat j$ and $\overrightarrow {B} = \cos \frac{\omega t}{2} \hat i + \sin \frac{\omega t}{2} \hat j$.Calculating the dot product:$\overrightarrow {A} \cdot \overrightarrow {B} = (\cos \omega t \hat i + \sin \omega t \hat j) \cdot (\cos \frac{\omega t}{2} \hat i + \sin \frac{\omega t}{2} \hat j)$$= \cos \omega t \cos \frac{\omega t}{2} + \sin \omega t \sin \frac{\omega t}{2}$Using the trigonometric identity $\cos(A - B) = \cos A \cos B + \sin A \sin B$:$\overrightarrow {A} \cdot \overrightarrow {B} = \cos(\omega t - \frac{\omega t}{2}) = \cos(\frac{\omega t}{2})$Since the vectors are orthogonal,$\overrightarrow {A} \cdot \overrightarrow {B} = 0$,so:$\cos(\frac{\omega t}{2}) = 0$We know $\cos(\frac{\pi}{2}) = 0$,therefore:$\frac{\omega t}{2} = \frac{\pi}{2}$$t = \frac{\pi}{\omega}$.

If vectors $\overrightarrow {A} = \cos\omega t\hat i + \sin\omega t\hat j$ and $\overrightarrow {B} = \cos\frac{\omega t}{2}\hat i + \sin\frac{\omega t}{2}\hat j$ are functions of time,then the value of $t$ at which they are orthogonal to each other is:

Similar Questions

The two vectors have magnitudes $3$ and $5$. If the angle between them is $60^o$,then the dot product of the two vectors will be:

Let $\vec A = (\hat i + \hat j)$ and $\vec B = (2\hat i - \hat j)$. The magnitude of a coplanar vector $\vec C$ such that $\vec A \cdot \vec C = \vec B \cdot \vec C = \vec A \cdot \vec B$ is given by

If $\overrightarrow{A} = (2\hat{i} + 3\hat{j} - \hat{k}) \; m$ and $\overrightarrow{B} = (\hat{i} + 2\hat{j} + 2\hat{k}) \; m$,the magnitude of the component of vector $\overrightarrow{A}$ along vector $\overrightarrow{B}$ will be $...... \; m$.

If two vectors $A$ and $B$ are mutually perpendicular,then the component of $A-B$ along the direction of $A+B$ is

Let $\vec{A} = 2 \hat{i} - 3 \hat{j} + 4 \hat{k}$ and $\vec{B} = 4 \hat{i} + \hat{j} + 2 \hat{k}$,then $|\vec{A} \times \vec{B}|$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module