MCQ
It is found that $|\text{A}+\text{B}|=|\text{A}|.$ This necessarily implies,
- A$\text{B}=0$
- BA, B are antiparallel.
- CA, B are perpendicular.
- D$\text{A.B}≤0$
✓
Answer
- $(\vec{\text{A}}\times\vec{\text{B}})\times\vec{\text{C}}$ is not zero unless $\vec{\text{B}},\ \vec{\text{C}}$ are parallel.
Explanation:
$|\vec{\text{A}}+\vec{\text{B}}|=|\vec{\text{A}}|$
Applying dot product,
$|\vec{\text{A}}+\vec{\text{B}}|.|\vec{\text{A}}+\vec{\text{B}}|=|\vec{\text{A}}|.|\vec{\text{A}}|$
$\Rightarrow\ |\vec{\text{A}}|^2+2\vec{\text{A}}.\vec{\text{B}}+|\vec{\text{B}}|^2=|\text{A}|^2$
$\Rightarrow\ |\vec{\text{B}}|^2=-2\vec{\text{A}}.\vec{\text{B}}$
$\Rightarrow\ |\vec{\text{B}}|=0$
Therefore, option (a) is correct.
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