Question

the resultant of two vectors P and Q is perpendicular to P and its magnitude is half of that of Q. Whats the angle between P and Q?

Answer 1

Answer:

The angle between P and Q is $150^0$

Explanation:

Given,

Resultant of two vectors P and Q is perpendicular to P.

The magnitude of the resultant vectors of P and Q is half of Q

We know, that the resultant vector 'R' of two vectors P and Q is given by the formula

R = $\sqrt{P^2 + Q^2 + 2PQCos\theta}$ ----------------(1)

where $\theta$ is the angle between the two vectors

Also, if $\alpha$ is the angle between P and R then, we have

$tan{\alpha} = \frac{QSin{\theta}}{P+ QCos\theta}$

Given that,

R is perpendicular to P, then we have $\alpha = 90$

$tan{90} = \frac{QSin{\theta}}{P+ QCos\theta}$

$\frac{1}{0} = \frac{QSin{\theta}}{P+ QCos\theta}$

$P + Q Cos\theta = 0$

$Cos \theta = \frac{-P}{Q}$ --------------(2)

Also, given R = $\frac{Q}{2}$

Substituting the value of 'R' and $Cos\theta$  in equation (1) we get

$\frac{Q}{2}$ = $\sqrt{P^2 + Q^2 + 2PQ ( -\frac{P}{Q}) }$

$\frac{Q}{2} = \sqrt{P^2 + Q^2 -2P^2}$

$\frac{Q}{2} = \sqrt{ Q^2 -P^2}$

$\frac{Q^2}{4} = Q^2 -P^2\\P^2 = Q^2 - \frac{Q^2}{4} \\P^2 = \frac{3Q^2}{4} \\P = \frac{\sqrt{3}}{2} Q$

$\frac{P }{Q} = \frac{\sqrt{3} }{2}$

Substitute the value  $\frac{P}{Q}$ in equation(2)

$Cos \theta = \frac{-\sqrt{3} }{2}$

$\theta = 150^0$

Hence angle between P and Q is $150^0$