The resultant of two vectors A and B is perpendicular to A and the...

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:

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

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

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

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

The magnitude of the resultant is given by:

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

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

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

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

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

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

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

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

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

$\theta = 135^\circ$

Answer: $135^\circ$