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

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

Question

The formula for $|\vec{A}+\vec{B}|^{2}$ is,

$|\vec{A}+\vec{B}|^{2}=|\vec{A}|^{2}+|\vec{B}|^{2}+2 \vec{A} \cdot \vec{B}$

$=A+B+2 A B \cos 	heta$

Vedclass · Accepted Answer

$90$

Vedclass · Answer

The formula for $|\vec{A}+\vec{B}|^{2}$ is,

$|\vec{A}+\vec{B}|^{2}=|\vec{A}|^{2}+|\vec{B}|^{2}+2 \vec{A} \cdot \vec{B}$

$=A+B+2 A B \cos 	heta$ And The formula for $|\vec{A}-\vec{B}|^{2}$ is,

$|\vec{A}-\vec{B}|^{2}=|\vec{A}|^{2}+|\vec{B}|^{2}-2 \vec{A} \cdot \vec{B}$

$=A+B-2 A B \cos 	heta$

It is given that,

$|\vec{A}+\vec{B}|^{2}=|\vec{A}-\vec{B}|^{2}$

$A+B+2 A B \cos 	heta=A+B-2 A B \cos 	heta$

$4 A B \cos 	heta=0$

$\cos 	heta=0$

$	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

Similar Questions

Two forces are such that the sum of their magnitudes is $18 \,N$ and their resultant is perpendicular to the smaller force and magnitude of resultant is $12\, N$. Then the magnitudes of the forces are

Figure shows three vectors $p , q$ and $r$, where $C$ is the mid point of $A B$. Then, which of the following relation is correct?

Mark the correct statement :-

Two forces $3\,N$ and $2\, N$ are at an angle $\theta$ such that the resultant is $R$. The first force is now increased to $ 6\,N$ and the resultant become $2R$. The value of is ....... $^o$

Given that $\overrightarrow A + \overrightarrow B = \overrightarrow C $and that $\overrightarrow C $ is $ \bot $ to $\overrightarrow A $. Further if $|\overrightarrow A |\, = \,|\overrightarrow C |,$then what is the angle between $\overrightarrow A $ and $\overrightarrow B $

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