If |vec{A}+vec{B}|=|vec{A}-vec{B}|,then the a | vedclass

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

Question

(C) Given the condition: $|\vec{A}+\vec{B}| = |\vec{A}-\vec{B}|$.Squaring both sides,we get: $|\vec{A}+\vec{B}|^2 = |\vec{A}-\vec{B}|^2$.Using the pro

Vedclass · Accepted Answer

$90$

Vedclass · Answer

(C) Given the condition: $|\vec{A}+\vec{B}| = |\vec{A}-\vec{B}|$.Squaring both sides,we get: $|\vec{A}+\vec{B}|^2 = |\vec{A}-\vec{B}|^2$.Using the property $|\vec{V}|^2 = \vec{V} \cdot \vec{V}$,we expand the terms:$A^2 + B^2 + 2AB \cos 	heta = A^2 + B^2 - 2AB \cos 	heta$.Subtracting $A^2 + B^2$ from both sides:$2AB \cos 	heta = -2AB \cos 	heta$.$4AB \cos 	heta = 0$.Since $A$ and $B$ are non-zero vectors,$\cos 	heta = 0$.Therefore,$	heta = 90^{\circ}$.

If $|\vec{A}+\vec{B}|=|\vec{A}-\vec{B}|$,then the angle between vector $\vec{A}$ and $\vec{B}$ is: (in $^{\circ}$)

Similar Questions

If a unit vector is represented by $0.5\hat i + 0.8\hat j + c\hat k$,then the value of $c$ is

The vector sum of two forces $\vec{A}$ and $\vec{B}$ is perpendicular to their vector difference. Hence,the forces $\vec{A}$ and $\vec{B}$ are

In a switching circuit, if the combination $(S_1 \land S_2)$ is connected in parallel to the combination $(S_1' \land S_2')$, then the room is lit only when

Vector $\overrightarrow{A}$ makes equal angles with $x, y,$ and $z$ axes. The value of its components (in terms of the magnitude of $\overrightarrow{A}$) will be:

If $\vec{A}$ and $\vec{B}$ are two non-zero vectors such that $|\vec{A} + \vec{B}| = \frac{|\vec{A} - \vec{B}|}{2}$ and $|\vec{A}| = 2|\vec{B}|$,then the angle between $\vec{A}$ and $\vec{B}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module