Physics - Vectors Question with Solution | TestHub

Let the angle between $\vec{A}$ and $\vec{B}$ be $\theta$.

The resultant of $\vec{A}$ and $\vec{B}$ is $\vec{C}=\vec{A}+\vec{B}$.

The square of 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 is $\vec{R}=\vec{A}+2\vec{B}$.

Since $\vec{R}$ is perpendicular to $\vec{A}$, their dot product is 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 in the equation for $C^2$:

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

$C^2=B^2$

$C=B$

Answer: B