# The resultant of two vectors P and Q is R.If Q is doubled then the new resultant vector is perpendicular to P.Then R is equal to?

Sep 21, 2017

Let the angle between two vectors $P \mathmr{and} Q$ be $\alpha$ and their resultant is $R$

So we can write

${R}^{2} = {P}^{2} + {Q}^{2} + 2 P Q \cos \alpha \ldots \ldots \left[1\right]$

When Q is doubled then let the resultant vector be ${R}_{1}$, So we can write

${R}_{1}^{2} = {P}^{2} + 4 {Q}^{2} + 4 P Q \cos \alpha \ldots \ldots \left[2\right]$

Again by the given condition ${R}_{1}$ is perpendicular to $P$

So $4 {Q}^{2} = {P}^{2} + {R}_{1}^{2.} \ldots . . \left[3\right]$

Combining [2] and [3] we get

${R}_{1}^{2} = {P}^{2} + {P}^{2} + {R}_{1}^{2} + 4 P Q \cos \alpha$

$\implies 2 P Q \cos \alpha = - {P}^{2.} \ldots . . \left[4\right]$

combining [1] and [4] we get

${R}^{2} = {P}^{2} + {Q}^{2} - {P}^{2}$

$\implies R = Q$