Which of the following is the unit vector per | vedclass

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

Question

(C) The cross product of two vectors $\vec{A}$ and $\vec{B}$ is defined as $\vec{A} 	imes \vec{B} = AB \sin 	heta \; \hat{n}$,where $\hat{n}$ is the

Vedclass · Accepted Answer

$\frac{\vec{A} 	imes \vec{B}}{AB \sin 	heta}$

Vedclass · Answer

(C) The cross product of two vectors $\vec{A}$ and $\vec{B}$ is defined as $\vec{A} 	imes \vec{B} = AB \sin 	heta \; \hat{n}$,where $\hat{n}$ is the unit vector perpendicular to both $\vec{A}$ and $\vec{B}$.From this definition,we can isolate the unit vector $\hat{n}$ as:$\hat{n} = \frac{\vec{A} 	imes \vec{B}}{AB \sin 	heta}$.Since $|\vec{A}| = A$ and $|\vec{B}| = B$,the expression becomes $\frac{\vec{A} 	imes \vec{B}}{|\vec{A}| |\vec{B}| \sin 	heta}$.Thus,option $C$ is correct.

Which of the following is the unit vector perpendicular to $\vec{A}$ and $\vec{B}$?

Similar Questions

$A$ vector $\vec{A}$ points vertically upward and $\vec{B}$ points towards north. The vector product $\vec{A} \times \vec{B}$ is

For any two vectors $\vec{A}$ and $\vec{B}$,if $\vec{A} \cdot \vec{B} = |\vec{A} \times \vec{B}|$,the magnitude of $(\vec{A} + \vec{B})$ is: $(\tan \frac{\pi}{4} = 1, \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}})$

Two position vectors are given by $\overrightarrow{r_1} = (1, 1, 1)$ and $\overrightarrow{r_2} = (1, -1, 1)$. Find the unit vector in the direction of $\overrightarrow{r_1} \times \overrightarrow{r_2}$.

The area of the parallelogram determined by vectors $A = 2\hat{i} + \hat{j} - 3\hat{k}$ and $B = 12\hat{j} - 2\hat{k}$ is approximately:

Show that the magnitude of a vector is equal to the square root of the scalar product of the vector with itself.

Vedclass Test Series

Exam Paper Generator

Online Exam Module