MCQ
For two vectors $A$ and $B, |\text{A} + \text{B}| = |\text{A} - \text{B}|$ is always true when
- A$|\text{A}|=|\text{B}|\neq0$
- B$\text{A}\bot\text{B}$
- C$|A|=|B|\neq 0$ and $A$ and $B$ are parallel or anti parallel
- ✓$B$ and $C$
✓
Answer
Correct option: D.
$B$ and $C$
According to the problem, $|\vec{\text{A}}+\vec{\text{B}}|=|\vec{\text{A}}-\vec{\text{B}}|$
$\Rightarrow\ \sqrt{|\vec{\text{A}}|^2+|\vec{\text{B}}|^2+2|\vec{\text{A}}||\vec{\text{B}}|\cos\theta}\\=\sqrt{|\vec{\text{A}}|^2+|\vec{\text{B}}|^2-2|\vec{\text{A}}||\vec{\text{B}}|\cos\theta}$
$\Rightarrow\ |\vec{\text{A}}|^2+|\vec{\text{B}}|^2+2|\vec{\text{A}}||\vec{\text{B}}|\cos\theta\\=|\vec{\text{A}}|^2+|\vec{\text{B}}|^2-2|\vec{\text{A}}||\vec{\text{B}}|\cos\theta$
$\Rightarrow\ 4|\vec{\text{A}}||\vec{\text{B}}|\cos\theta=0$
$\Rightarrow\ |\vec{\text{A}}||\vec{\text{B}}|\cos\theta=0$
$|\vec{\text{A}}|=0$ or $|\vec{\text{B}}|=0$ or $\cos\theta=0$
i.e. $\theta=90^\circ$
When $\theta=90^\circ,$ we can say that $\vec{\text{A}}\bot\vec{\text{B}}.$
Hence options $(B)$ and $(C)$ are correct.
$\Rightarrow\ \sqrt{|\vec{\text{A}}|^2+|\vec{\text{B}}|^2+2|\vec{\text{A}}||\vec{\text{B}}|\cos\theta}\\=\sqrt{|\vec{\text{A}}|^2+|\vec{\text{B}}|^2-2|\vec{\text{A}}||\vec{\text{B}}|\cos\theta}$
$\Rightarrow\ |\vec{\text{A}}|^2+|\vec{\text{B}}|^2+2|\vec{\text{A}}||\vec{\text{B}}|\cos\theta\\=|\vec{\text{A}}|^2+|\vec{\text{B}}|^2-2|\vec{\text{A}}||\vec{\text{B}}|\cos\theta$
$\Rightarrow\ 4|\vec{\text{A}}||\vec{\text{B}}|\cos\theta=0$
$\Rightarrow\ |\vec{\text{A}}||\vec{\text{B}}|\cos\theta=0$
$|\vec{\text{A}}|=0$ or $|\vec{\text{B}}|=0$ or $\cos\theta=0$
i.e. $\theta=90^\circ$
When $\theta=90^\circ,$ we can say that $\vec{\text{A}}\bot\vec{\text{B}}.$
Hence options $(B)$ and $(C)$ are correct.
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