The resultant of u¯ and v¯ is perpendicular to u¯ and has a magnitude equal to half of the magnitude of v¯. The angle between u¯&v¯ is

Given α=90∘&R=12ϑ;θ=?Tan⁡α=ϑsin⁡θu+ϑcos⁡θ⇒u+ϑcos⁡θ=0⇒cos⁡θ=−uϑ ; R2=u2+ϑ2+2uϑcos⁡θϑ24=ϑ2−u2⇒u2=3ϑ24uϑ=32;cos⁡θ=−32⇒θ=150∘

The resultant of u¯ and v¯ is perpendicular to u¯ and has a magnitude equal to half of the magnitude of v¯. The angle between u¯&v¯ is

Given α=90∘&R=12ϑ;θ=?Tan⁡α=ϑsin⁡θu+ϑcos⁡θ⇒u+ϑcos⁡θ=0⇒cos⁡θ=−uϑ ; R2=u2+ϑ2+2uϑcos⁡θϑ24=ϑ2−u2⇒u2=3ϑ24uϑ=32;cos⁡θ=−32⇒θ=150∘

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The resultant of $\bar{u} and \bar{v}$ is perpendicular to $\bar{u}$ and has a magnitude equal to half of the magnitude of $\bar{v}$ . The angle between $\bar{u} & \bar{v} is$

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$120^{\circ}$

$60^{\circ}$

$150^{\circ}$

$90^{\circ}$

answer is B.

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Detailed Solution

$Given α = 90^{\circ} & R = \frac{1}{2} ϑ; θ = ?$

$Tan α = \frac{ϑ \sin θ}{u + ϑ \cos θ} \Rightarrow u + ϑ \cos θ = 0$

$\Rightarrow \cos θ = - \frac{u}{ϑ}; R^{2} = u^{2} + ϑ^{2} + 2 u ϑ \cos θ$

$\frac{ϑ^{2}}{4} = ϑ^{2} - u^{2} \Rightarrow u^{2} = \frac{3 ϑ^{2}}{4}$

$\frac{u}{ϑ} = \frac{\sqrt{3}}{2}; \cos θ = - \frac{\sqrt{3}}{2} \Rightarrow θ = 150^{\circ}$

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The resultant of u¯ and v¯ is perpendicular to u¯ and has a magnitude equal to half of the magnitude of v¯. The angle between u¯&v¯ is

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The resultant of $\bar{u} and \bar{v}$ is perpendicular to $\bar{u}$ and has a magnitude equal to half of the magnitude of $\bar{v}$ . The angle between $\bar{u} & \bar{v} is$