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

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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.
D
$(\vec{\text{A}}\times\vec{\text{B}}).\vec{\text{C}}= |\text{A}||\text{B}||\text{C}|\rightarrow\text{C}^2=\text{A}^2+\text{B}^2$

✓

Answer

$(\vec{\text{A}}\times\vec{\text{B}})\times\vec{\text{C}}$ is not zero unless $\vec{\text{B}},\ \vec{\text{C}}$ are parallel.

If A, B, C define a plane, $(\vec{\text{A}}\times\vec{\text{B}})\times\vec{\text{C}}$$$ is in that plane.

Explanation:

$\vec{\text{A}}+\vec{\text{B}}+\vec{\text{C}}=0$

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.

$\vec{\text{B}}\times(\vec{\text{A}}+\vec{\text{B}}+\vec{\text{C}})=\vec{\text{B}}\times0=0$

$\vec{\text{B}}\times\vec{\text{A}}+\vec{\text{B}}\times\vec{\text{B}}+\vec{\text{B}}\times\vec{\text{C}}=0$

$\vec{\text{B}}\times\vec{\text{A}}+0+\vec{\text{B}}\times\vec{\text{C}}=0$

$\vec{\text{B}}\times\vec{\text{A}}=-\vec{\text{B}}\times\vec{\text{C}}$

$\vec{\text{A}}\times\vec{\text{B}}=\vec{\text{B}}\times\vec{\text{C}}\ \dots(\text{i})$

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.

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

$(\vec{\text{A}}\times\vec{\text{B}})=\vec{\text{X}}$

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 $\vec{\text{X}}\text{ and }\vec{\text{C}}$ which again become in the plane of $\vec{\text{A}},\ \vec{\text{B}},\ \vec{\text{C}}$ but perpendicular to the plane of $\vec{\text{X}}\text{ and }\vec{\text{C}}.$ Hence option (c) is also verified.