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3-1.Vectors
hard
સદિશ $\vec{A}$ અને $\vec{B}$ એવા છે કે જેથી $|\vec{A}+\vec{B}|=|\vec{A}-\vec{B}|$ થાય. બે સદિશ વચ્ચેનો ખૂણો કેટલો હશે?
A
$60$
B
$75$
C
$45$
D
$90$
(AIIMS-2019) (AIPMT-1996) (AIPMT-2006) (AIPMT-1991) (AIIMS-2016)
Solution
The formula for $|\vec{A}+\vec{B}|^{2}$ is,
$|\vec{A}+\vec{B}|^{2}=|\vec{A}|^{2}+|\vec{B}|^{2}+2 \vec{A} \cdot \vec{B}$
$=A+B+2 A B \cos \theta$ And The formula for $|\vec{A}-\vec{B}|^{2}$ is,
$|\vec{A}-\vec{B}|^{2}=|\vec{A}|^{2}+|\vec{B}|^{2}-2 \vec{A} \cdot \vec{B}$
$=A+B-2 A B \cos \theta$
It is given that,
$|\vec{A}+\vec{B}|^{2}=|\vec{A}-\vec{B}|^{2}$
$A+B+2 A B \cos \theta=A+B-2 A B \cos \theta$
$4 A B \cos \theta=0$
$\cos \theta=0$
$\theta=90^{\circ}$
Standard 11
Physics
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