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3-1.Vectors
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The resultant of two vectors $A$ and $B$ is perpendicular to the vector $A$ and its magnitude is equal to half the magnitude of vector $B$. The angle between $A$ and $B$ is ....... $^o$
A$120$
B$150$
C$135$
DNone of these
Solution
(b) $\frac{B}{2} = \sqrt {{A^2} + {B^2} + 2AB\;\cos \theta } $…$(i)$
$\tan 90^\circ = \frac{{B\sin \theta }}{{A + B\cos \theta }} \Rightarrow A + B\cos \theta = 0$
$\cos \theta = – \frac{A}{B}$
Hence, from $(i)$ $\frac{{{B^2}}}{4} = {A^2} + {B^2} – 2{A^2} \Rightarrow A = \sqrt 3 \frac{B}{2}$
$⇒ \cos \theta = – \frac{A}{B} = – \frac{{\sqrt 3 }}{2}$
$⇒ \theta = 150^\circ $
$\tan 90^\circ = \frac{{B\sin \theta }}{{A + B\cos \theta }} \Rightarrow A + B\cos \theta = 0$
$\cos \theta = – \frac{A}{B}$
Hence, from $(i)$ $\frac{{{B^2}}}{4} = {A^2} + {B^2} – 2{A^2} \Rightarrow A = \sqrt 3 \frac{B}{2}$
$⇒ \cos \theta = – \frac{A}{B} = – \frac{{\sqrt 3 }}{2}$
$⇒ \theta = 150^\circ $
Standard 11
Physics
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