The resultant of two vectors u- and v- is perpendicular to the vector u- and its magnitude is equal to half of the magnitude of vector v- The angle between u- and v- is- -

The correct option is B 150^∘ According to question: (u⃗+v⃗).u⃗=0 and u⃗+v⃗=v/2 ⇒ u^2+u⃗.v⃗=0 amp; u^2+v^2+2u⃗.v⃗=v^2/4 ⇒3/4v ...

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Byju's Answer

Standard XII

Physics

Angular Velocity

The resultant...

Question

The resultant of two vectors $\to u$ and $\to v$ is perpendicular to the vector $\to u$ and its magnitude is equal to half of the magnitude of vector $\to v$ . The angle between $\to u$ and $\to v$ is......... .

150^{\circ}

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180^{o}

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90^{o}

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60^{o}

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Solution

The correct option is B $150^{\circ}$
According to question:
$(\to u + \to v) . \to u = 0$ and $| \begin{matrix} \to u + \to v \end{matrix} | = \frac{v}{2}$
$\Rightarrow u^{2} + \to u . \to v = 0$ & $u^{2} + v^{2} + 2 \to u . \to v = \frac{v^{2}}{4}$
$\Rightarrow \frac{3}{4} v^{2} = u^{2} \to c o s θ = - \frac{\sqrt{3}}{2} \Rightarrow θ = 150^{\circ}$

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The resultant of two vectors →u and →v is perpendicular to the vector →u and its magnitude is equal to half of the magnitude of vector →v. The angle between →u and →v is......... .

The correct option is B 150∘According to question:(→u+→v).→u=0 and |→u+→v|=v2⇒u2+→u.→v=0 & u2+v2+2→u.→v=v24⇒34v2=u2→cosθ=−√32⇒θ=150∘

The resultant of two vectors $\to u$ and $\to v$ is perpendicular to the vector $\to u$ and its magnitude is equal to half of the magnitude of vector $\to v$ . The angle between $\to u$ and $\to v$ is......... .