The resultant of two vectors P- and Q- is R- lf the magnitude of Q- is doubled- the new resultant becomes perpendicular to P- then magnitude of R- is -

The correct option is A Q Resultant of two vector P⃗ and Q⃗ is R⃗=√(P^2+Q^2+2PQ cosθ) . #160;Where θ is the angle between the two ...

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Byju's Answer

Standard XII

Physics

Vector Component

The resultant...

Question

The resultant of two vectors $\to P$ and $\to Q$ is $\to R$ . lf the magnitude of $\to Q$ is doubled, the new resultant becomes perpendicular to $\to P$ , then magnitude of $\to R$ is :

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

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

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Q

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\frac{P}{Q}

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Solution

The correct option is A $Q$
Resultant of two vector $\to P$ and $\to Q$ is $\to R = \sqrt{P^{2} + Q^{2} + 2 P Q cos θ}$ .
Where $θ$ is the angle between the two vectors.
As magnitude of $\to Q$ is doubled the angle between the new resultant is $90^{o}$
Thus we get $t a n 90^{o} = \frac{2 Q s i n θ}{P + 2 Q c o s θ}$
As $t a n 90^{o} = \infty$ we have $P + 2 Q c o s θ = 0$

$P = - 2 Q c o s θ$

Now, $R^{2} = P^{2} + Q^{2} + 2 P Q \frac{- P}{2 Q}$

$R = Q$

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

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