Question

Two forces P and Q acting at a point are such that if P is reversed, the direction of the resultant is turned through 90o. Then

a) P = Q

b) P = 2Q

c) P = Q/2

d) No relation between P and Q

Answer 1

Hi_________
it just simle question which is based upon vector.
option (a) is correct
P=Q
hope it help u

Answer 2

Answer:

a) P = Q

Explanation:

Let

$P=a \hat{\imath}+b \hat{\jmath}$

$Q=c \hat{\imath}+d \hat{\jmath}$

Magnitude will be,

$|P|=\sqrt{a^{2}+b^{2}}$

$|Q|=\sqrt{c^{2}+d^{2}}$

If have 90 degree angle,

$R=R_{1}+R_{2}$

$R=Q+P=(a+c) \hat{\imath}+(b+d) \hat{\jmath} \rightarrow(1)$

After p is reversed,

$R=Q-P=(c-a) \hat{\imath}+(d-b) \hat{\jmath} \rightarrow(2)$

The equations (1) and (2) are perpendicular to each other,

So,

$A . B=|A||B| \cos \theta$

And

$\theta=90^{\circ}$

Since,

$\cos 90=0$

Therefore, $A.B = 0$

$\Rightarrow[(a+c) \hat{\imath}+(b+d) \hat{\jmath}] \cdot[(c-a) \hat{\imath}+(d-b) \hat{\jmath}]=0$

On multiplying, we get,

$\Rightarrow a c-a^{2}+c^{2}-a c+b d-b^{2}+d^{2}-b d=0$

$\Rightarrow c^{2}-a^{2}-b^{2}+d^{2}=0$

On bringing negative and positive variables on either sides, we get.

$\Rightarrow c^{2}+d^{2}=a^{2}+b^{2}$

$\Rightarrow|Q|=|P|$

Therefore, the correct option is (a)