Motion in a Plane — Physics STD 11 Science — Question

1. $(\vec{\text{A}}\times\vec{\text{B}})\times\vec{\text{C}}$ 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.