# the resultant of two vectors P and Q is R. if Q is doubled, the new resultant is perpendicular to P. Then R equals(a) P(b) (P+Q)( c) Q(d) (P-Q) please explain how to solve

Saurabh Kumar
8 years ago
Surya singh
36 Points
5 years ago
For first scene when resultant of P and Q is R
Vector R=Vector P +Vector Q
R=P2 +Q2+2PQCos$\Theta$….................$equation 1$
Then after doubling the magnitude of Q the new resultant is perpendicular to P..That means dot product of P and new resultant(R’) will be zero.
R’ * P=0...................equation 2
(where R’=P+2Q)
by putting value to equation 2
we get Cos$\Theta$ = -P/2Q............equation 3
So from equation 1 and 3,we get
R=Q

3 years ago
Hello student

By law of vector addition, the resultant vector can be given as
R = P + Q

Now when Q is doubled the new resultant vector can be given as
R’ = P + 2Q
since, R’ is perpendicular to P
hence there dot product should be zero
or, R’ . P = 0
or, ( P + 2Q ).P = |P|2 + 2Q.P = 0
or, 2P.Q = -|P|2

Now,  |R|2 = |P|2 + |Q|2 + 2P.Q
=  |P|2 + |Q|2 – |P|2
=  |Q|2
or, R = Q

Hope it helps
Regards
Kushagra
Yash Chourasiya
3 years ago
Dear Student

Let the angle between two vectorsP and Qbeαand their resultant isR

So we can write

R2 = P2 + Q2 + 2PQcosα......[1]

When Q is doubled then let the resultant vector beR1, So we can write

R12= P2 + 4Q2 + 4PQcosα......[2]

Again by the given conditionR1is perpendicular toP

So4Q2 = P2 + R12......[3]

Combining [2] and [3] we get

R12 = P2 + P2 + R12 + 4PQcosα

⇒ 2PQcosα = −P2......[4]

combining [1] and [4] we get

R2 = P2 + Q2 − P2

⇒ R = Q