The magnitude of the sum of the two vectors $\overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{B}}$…

The magnitude of the sum of the two vectors $\overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{B}}$ is equal to the magnitude of the difference of two vectors $\overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{B}}$. The angle between $\overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{B}}$ is
  1. $30^{\circ}$
  2. $45^{\circ}$
  3. $90^{\circ}$
  4. $180^{\circ}$

Solution

Let the two vectors be $\vec{A}$ and $\vec{B}$ with magnitudes $A$ and $B$ respectively. Then magnitude of their sum is given by: $|\vec{A}+\vec{B}|=\sqrt{A^{2}+B^{2}+2 A B \cos \theta} \rightarrow(1)(\theta=$ angle between the vectors $)$ Magnitude of their difference is given by: $|\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}}|=\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}-2 \mathrm{AB} \cos \theta} \rightarrow(2)$ As $|\vec{A}+\vec{B}|=|\vec{A}-\vec{B}|$ $\Rightarrow \mathrm{A}^{2}+\mathrm{B}^{2}+2 \mathrm{AB} \cos \theta=\mathrm{A}^{2}+\mathrm{B}^{2}-2 \mathrm{AB} \cos \theta$ $\Rightarrow 4 \mathrm{AB} \cos \theta=0$ or $\cos \theta=0$ $\Rightarrow \theta=90^{\circ}$.

Asked in: MHT CET 2020 (20 Oct Shift 1)

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