Three vectors A , B and C add up to zero. Find which is false.

MCQ

Three vectors $\vec{\text{A}},\ \vec{\text{B}}$ and $\vec{\text{C}}$ add up to zero. Find which is false.

A
$(\vec{\text{A}}\times\vec{\text{B}})\times\vec{\text{C}}$ is not zero unless $\vec{\text{B}},\ \vec{\text{C}}$ are parallel.
B
$(\vec{\text{A}}\times\vec{\text{B}}).\vec{\text{C}}$ is not zero unless $\vec{\text{B}},\ \vec{\text{C}}$ are parallel.
C
If $A, B, C$ define a plane, $(\vec{\text{A}}\times\vec{\text{B}})\times\vec{\text{C}}$ is in that plane.
✓
Both $A$ and $C$

✓

Answer

Correct option: D.

Both $A$ and $C$

$\vec{\text{A}}+\vec{\text{B}}+\vec{\text{C}}=0$
So $\vec{\text{A}},\ \vec{\text{B}}$ and $\vec{\text{C}}$ are in a plane and can be represented bt the three sides of a triangle taken in order.
$a. \vec{\text{B}}\times(\vec{\text{A}}+\vec{\text{B}}+\vec{\text{C}})=\vec{\text{B}}\times0=0$
$\vec{\text{B}}\times\vec{\text{A}}+\vec{\text{B}}\times\vec{\text{B}}+\vec{\text{B}}\times\vec{\text{C}}=0$
$\vec{\text{B}}\times\vec{\text{A}}+0+\vec{\text{B}}\times\vec{\text{C}}=0$
$\vec{\text{B}}\times\vec{\text{A}}=-\vec{\text{B}}\times\vec{\text{C}}$
$\vec{\text{A}}\times\vec{\text{B}}=\vec{\text{B}}\times\vec{\text{C}}\ \dots(\text{i})$
Or $(\vec{\text{A}}\times\vec{\text{B}})\times\vec{\text{C}}=(\vec{\text{B}}\times\vec{\text{C}})\times\vec{\text{C}}$
$\therefore$ It cannot be zero
$\vec{\text{B}}\times\vec{\text{C}}$ will be zero if $\vec{\text{B}}\times\vec{\text{C}}$ are parallel or antiparallel.
i.e, $(\vec{\text{A}}\times\vec{\text{B}})\times\vec{\text{C}}=[\text{BC}\sin0^\circ]\times\vec{\text{C}}$
$(\vec{\text{A}}\times\vec{\text{B}})\times\vec{\text{C}}=0$ only if $\vec{\text{B}}||\vec{\text{C}}$
Hence option $(a)$ is verified.
$b. (\vec{\text{A}}\times\vec{\text{B}})=\vec{\text{B}}\times\vec{\text{C}} [$from $(i)]$
$(\vec{\text{A}}\times\vec{\text{B}}).\vec{\text{C}}=(\vec{\text{B}}\times\vec{\text{C}}).\vec{\text{C}}$
if $\vec{\text{B}}||\vec{\text{C}}$
$\vec{\text{B}}\times\vec{\text{C}}=\text{BC}\sin0^\circ=0$
$\therefore\ (\vec{\text{A}}\times\vec{\text{B}}).\vec{\text{C}}=0\text{ IF }\vec{\text{B}}||\vec{\text{C}}$
So option $(b)$ is not verified.
$c. (\vec{\text{A}}\times\vec{\text{B}})=\vec{\text{X}}$
The direction of $x$ is perpendicular to both planes containing $A$ and $B.$
$(\vec{\text{A}}\times\vec{\text{B}})\times\vec{\text{C}}=\vec{\text{X}}\times\vec{\text{C}}=\vec{\text{Y}}$
The direction of $\vec{\text{Y}}$ is perpendicular to the plane of $\vec{\text{X}}$ and $\vec{\text{C}}$ which again become in the plane of $\vec{\text{A}},\ \vec{\text{B}},\ \vec{\text{C}}$ but perpendicular to the plane of $\vec{\text{X}}$ and $\vec{\text{C}}.$
Hence option $(c)$ is also verified.
$d. |\text{A}|^2+|\text{B}|^2=|\text{C}|^2$ given
It shows that angle between $\vec{\text{A}}$ and $\vec{\text{B}}$ is $90^\circ$
$=|\text{A}||\text{B}||\text{C}|\cos\theta\neq|\text{A}||\text{B}||\text{C}|$
$(\vec{\text{A}}\times\vec{\text{B}}).\vec{\text{C}}=|\text{A}||\text{B}||\text{C}|\cos\theta$
Does not verified option $(c).$