First slide

Vectors

Question

The resultant of two vectors P and Q is R. If Q is doubled, the new resultant is perpendicular to P. Then, R equal to

Moderate

A

P
B

(P+Q)
C

Q
D

(P $-$ Q)

Solution

We have, 2Q sin $θ$ = P, as R $⊥$ P
Now, $R = \sqrt{P^{2} + Q^{2} + 2 P Q \cos (90 ° + θ)}$
$= \sqrt{P^{2} + Q^{2} - 2 P Q \sin θ}$
$= \sqrt{(2 Q \sin θ)^{2} + Q^{2} - 2 (2 Q \sin θ) Q \sin θ}$
$= \sqrt{4 Q^{2} \sin^{2} θ + Q^{2} - 4 Q^{2} \sin^{2} θ} = Q$

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