Question

When two vector of magnitudes P and Q are inclined at on angle theta. THE magnitude of their resultant is 2P. When the inclination is changed to 180-theta, the magnitude of their resultant is P. FIND THE RATIO OF p and q

Answer 1

$\huge{\underline{\underline{\red{Answer}}}}$

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$\underline{\mathbb{\pink{GIVEN}}}$

1st case

• Two vectors P and Q
• Angle between them is $\theta$
• Magnitude of resultant is P

2nd case

• The angle is 180-$\theta$
• The resultant is P

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CASE I

$2p = \sqrt{ {p}^{2} + {q}^{2} + 2pq \cos \theta }$

$4 {p}^{2} = {p}^{2} + {q}^{2} + 2pq \cos( \theta)$

$3 {p}^{2} = {q}^{2} + 2pq \cos \theta - - - - - - - i$

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CASE II

$p = \sqrt{ {p}^{2} + {q}^{2} + 2pq \cos(180 - \theta) } \\ p = \sqrt{ {p}^{2} + {q}^{2} - 2pq \cos( \theta) }$

${p}^{2} = {p}^{2} + {q}^{2} - 2pq \cos( \theta) - - - - - ii$

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From adding i and ii we get,

${q}^{2} + 2pq \cos( \theta) + {q}^{2} - 2pq \cos( \theta) = 3 {p}^{2} + 0 \\ 2{q}^{2} = 3 {p}^{2}$

$\frac{ {p}^{2} }{ {q}^{2} } = \frac{2}{3} \\ \frac{p}{q} = \frac{ \sqrt{2} }{ \sqrt{3} }$

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${\underline{\red{Answer\:is\: \frac{ \sqrt{2} }{ \sqrt{3} }}}}$