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

Question

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 (A)  P+Q  (B)  Q  (C)  P  (D)  (P+Q/2)

Tardigrade Admin · Accepted Answer

In first case R=√P2+Q2+2PQ cos θ ...(i) where θ is the angle between P and Q. In second case, (P+2Q).P=0 or PP+2(Q.P)=0 or P2+0=0 or P=0 Putting P=0 in Eq. (i), we get R=√0+Q2+0=Q

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

$P + Q$

$Q$

$P$

$\frac{P + Q}{2}$

$P - Q$

In first case
$R = P^{2} + Q^{2} + 2 PQ cos θ$ ...(i)
where $θ$ is the angle between P and Q. In second case,
$(P + 2 Q) . P = 0$ or $P P + 2 (Q . P) = 0$
or $P^{2} + 0 = 0$
or $P = 0$ Putting $P = 0$ in Eq. (i),
we get $R = 0 + Q^{2} + 0 = Q$

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

P+Q

Q

P

2P+Q​

P−Q

In first case R=P2+Q2+2PQcosθ​ ...(i) where θ is the angle between P and Q. In second case, (P+2Q​).P=0 or PP+2(Q​.P)=0 or P2+0=0 or P=0 Putting P=0 in Eq. (i), we get R=0+Q2+0​=Q

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

$P + Q$

$Q$

$P$

$\frac{P + Q}{2}$

$P - Q$

In first case
$R = P^{2} + Q^{2} + 2 PQ cos θ$ ...(i)
where $θ$ is the angle between P and Q. In second case,
$(P + 2 Q) . P = 0$ or $P P + 2 (Q . P) = 0$
or $P^{2} + 0 = 0$
or $P = 0$ Putting $P = 0$ in Eq. (i),
we get $R = 0 + Q^{2} + 0 = Q$