The vectors vec{A} and vec{B} are such that | | vedclass

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

Question

(D) The magnitude of the sum of two vectors is given by $|\vec{A}+\vec{B}|^2 = |\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}| \cos 	heta$.Similarly,

Vedclass · Accepted Answer

$90$

Vedclass · Answer

(D) The magnitude of the sum of two vectors is given by $|\vec{A}+\vec{B}|^2 = |\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}| \cos 	heta$.Similarly,the magnitude of the difference of two vectors is given by $|\vec{A}-\vec{B}|^2 = |\vec{A}|^2 + |\vec{B}|^2 - 2|\vec{A}||\vec{B}| \cos 	heta$.Given that $|\vec{A}+\vec{B}| = |\vec{A}-\vec{B}|$,we square both sides to get $|\vec{A}+\vec{B}|^2 = |\vec{A}-\vec{B}|^2$.Substituting the expressions,we have $|\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}| \cos 	heta = |\vec{A}|^2 + |\vec{B}|^2 - 2|\vec{A}||\vec{B}| \cos 	heta$.Canceling the common terms $|\vec{A}|^2$ and $|\vec{B}|^2$ from both sides,we get $2|\vec{A}||\vec{B}| \cos 	heta = -2|\vec{A}||\vec{B}| \cos 	heta$.Rearranging the terms,we get $4|\vec{A}||\vec{B}| \cos 	heta = 0$.Since the vectors are non-zero,$|\vec{A}| 
eq 0$ and $|\vec{B}| 
eq 0$,therefore $\cos 	heta = 0$.This implies $	heta = 90^{\circ}$.

The vectors $\vec{A}$ and $\vec{B}$ are such that $|\vec{A}+\vec{B}|=|\vec{A}-\vec{B}|$. The angle between the two vectors is: (in $^{\circ}$)

Similar Questions

For the given figure,which of the following relations is correct?

The $x$-component of the resultant of several vectors is:

The sum of the magnitudes of two vectors $\vec{A}$ and $\vec{B}$ is $8$,and the magnitude of their resultant is $4$. If the resultant vector is perpendicular to one of the vectors,then the magnitudes of the two vectors $\vec{A}$ and $\vec{B}$ are:

Find the resultant of three vectors $\overrightarrow{OA}$,$\overrightarrow{OB}$,and $\overrightarrow{OC}$ as shown in the figure. The radius of the circle is $R$.

If the magnitude of the resultant of two vectors of equal magnitude is equal to the magnitude of either vector,then find the angle between the two vectors. (in $^{\circ}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module