Physics - Vectors Question with Solution | TestHub

Given that the resultant of $\vec{A}$ and $\vec{B}$ is $\vec{C}$, we have:

$\vec{C}=\vec{A}+\vec{B}$

The magnitude of $\vec{C}$ is given by:

$C^2=A^2+B^2+2AB\cos \theta$

When the magnitude of $\vec{B}$ is doubled, the new resultant vector is $\vec{R}=\vec{A}+2\vec{B}$.

It is given that $\vec{R}$ is perpendicular to $\vec{A}$, so their dot product must be zero:

$\vec{R}\cdot \vec{A}=0$

$(\vec{A}+2\vec{B})\cdot \vec{A}=0$

$A^2+2\vec{A}\cdot \vec{B}=0$

$A^2+2AB\cos \theta =0$

$2AB\cos \theta =-A^2$

Substituting this value into the equation for $C^2$:

$C^2=A^2+B^2+(-A^2)$

$C^2=B^2$

$C=B$

Answer: $B$