Q192 / 200 · MHT CET - Medical 2026 (21 April Shift 1) · Physics
The resultant of two vectors $\vec{A}$ and $\vec{B}$ is perpendicular to $\vec{A}$ and the magnitude is $\dfrac{1}{\sqrt{2}}$ times that of $\vec{B}$ . The angle between $\vec{A}$ and $\vec{B}$ is
- A$45^\circ$
- Correct answerB$135^\circ$
- C$90^\circ$
- D$120^\circ$
Solution
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: $135^\circ$
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: $135^\circ$
Hence, the correct answer is option: 'B'.