Physics - Vectors Question with Solution | TestHub

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

The resultant vector is $\vec{R}=\vec{A}+\vec{B}$.

Since $\vec{R}$ is perpendicular to $\vec{A}$, their dot product is zero:

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

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

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

$\cos \theta =-\frac{A}{B}$

The magnitude of the resultant is given by:

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

Substituting $\cos \theta =-\frac{A}{B}$ into the equation:

$R^2=A^2+B^2+2AB\left(-\frac{A}{B}\right)$

$R^2=B^2-A^2$

It is given that $R=\frac{B}{\sqrt{2}}$, so $R^2=\frac{B^2}{2}$.

$\frac{B^2}{2}=B^2-A^2$

$A^2=\frac{B^2}{2}\Rightarrow A=\frac{B}{\sqrt{2}}$

Substituting the value of $A$ in the expression for $\cos \theta$:

$\cos \theta =-\frac{\frac{B}{\sqrt{2}}}{B}=-\frac{1}{\sqrt{2}}$

$\theta =135^{\circ }$

Answer: $135^{\circ }$