The resultant of two vectors P→ and Q→ is R→. If the magnitude of Q→ is doubled, the new resultant becomes perpendicular to P→, then the magnitude of R→ is

tan⁡90∘=2Qsin⁡θP+2Qcos⁡θ (∵magnitude is doubled)⇒∞=2Qsin⁡θP+2Qcos⁡θ⇒P+2qcos⁡θ=0R=P2+Q2+2PQcos⁡θ=Q2+P(P+2Qcos⁡θ)=Q2⇒R=Q

The resultant of two vectors P→ and Q→ is R→. If the magnitude of Q→ is doubled, the new resultant becomes perpendicular to P→, then the magnitude of R→ is

tan⁡90∘=2Qsin⁡θP+2Qcos⁡θ (∵magnitude is doubled)⇒∞=2Qsin⁡θP+2Qcos⁡θ⇒P+2qcos⁡θ=0R=P2+Q2+2PQcos⁡θ=Q2+P(P+2Qcos⁡θ)=Q2⇒R=Q

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The resultant of two vectors $\vec{P} and \vec{Q} is \vec{R}$ . If the magnitude of $\vec{Q}$ is doubled, the new resultant becomes perpendicular to $\vec{P}$ , then the magnitude of $\vec{R}$ is

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$\frac{P^{2} - Q}{2 PQ}$

$\frac{P + Q}{P - Q}$

$\frac{P}{Q}$

answer is C.

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Detailed Solution

$\tan 90^{\circ} = \frac{2 Qsin θ}{P + 2 Qcos θ}$ $(∵ magnitude is doubled)$
$\begin{array}{l} \Rightarrow \infty = \frac{2 Qsin θ}{P + 2 Qcos θ} \Rightarrow P + 2 qcos θ = 0 \\ R = \sqrt{P^{2} + Q^{2} + 2 PQcos θ} \\ = \sqrt{Q^{2} + P (P + 2 Qcos θ)} \\ = \sqrt{Q^{2}} \\ \Rightarrow R = Q \end{array}$