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Let $\overrightarrow a ,\overrightarrow b ,\overrightarrow c $ be unit vector such that $\overrightarrow a + \overrightarrow b + \overrightarrow c = 0$. Which of the following is correct ?

(A) $\overrightarrow a \times \overrightarrow b = \overrightarrow b \times \overrightarrow c = \overrightarrow c \times \overrightarrow a = \overrightarrow 0 $

(B) $\overrightarrow a \times \overrightarrow b = \overrightarrow b \times \overrightarrow c = \overrightarrow c \times \overrightarrow a \ne \overrightarrow 0 $

(C) $\overrightarrow a \times \overrightarrow b = \overrightarrow b \times \overrightarrow c = \overrightarrow a \times \overrightarrow c \ne \overrightarrow 0 $

(D) $\overrightarrow a \times \overrightarrow b ,\overrightarrow b \times \overrightarrow c ,\overrightarrow c \times \overrightarrow a $ are mutually perpendicular

The official answer is (B).

My approach is s follow

$\overrightarrow a + \overrightarrow b + \overrightarrow c = 0$

$\overrightarrow a \times \overrightarrow b + \overrightarrow b \times \overrightarrow b + \overrightarrow c \times \overrightarrow b = 0$

$\overrightarrow a \times \overrightarrow b = - \overrightarrow c \times \overrightarrow b = \overrightarrow b \times \overrightarrow c $

$\overrightarrow a \times \overrightarrow c + \overrightarrow b \times \overrightarrow c + \overrightarrow c \times \overrightarrow c = 0 \Rightarrow \overrightarrow b \times \overrightarrow c = \overrightarrow c \times \overrightarrow a $.

Now I am confused between option (A) and (B) if $\overrightarrow a ,\overrightarrow b ,\overrightarrow c $ are parallel to each other then option (A) is correct or option (B)

Vector being a unit vector plays any vital role in deciding between (A) and (B)

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    $\begingroup$ $c = -a -b$, so $c$ lies in the plane spanned by $a$ and $b$. The only possibility is that $a$, $b$ and $c$ are unit vectors in that plane and that $\angle ab = \angle bc = \angle ca = 120^\circ$. No two of them can be parallel. $\endgroup$
    – Rob Arthan
    Jan 31, 2021 at 14:38

3 Answers 3

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If they are all parallel to each other then $\vec{b} = k\vec{a} $, $\ \vec{c} = m\vec{a} $ which means that $(1+k+m)\vec{a} = \vec{0} $

We can't have $\vec{a} = \vec{0}$ since $\vec{a}$ is a unit vector. Similarly we notice that since all three are unit vectors then $k$ and $m$ must be either 1 or -1 so $k + m + 1 \neq 0$ which proves that you can't have three linearly dependent unit vectors that are parallel to each other (try to think of this geometrically - what would the sum of two parallel unit vectors on a plane be?)

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(A) would imply the vectors are pairwise parallel or antiparallel, so their sum is of length $1$ or $3$.

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If $\vec a, \vec b$ are parallel, they have the same direction.

Since they are both unit vectors, this implies $\vec a = \vec b$ and $\vec c = -2\vec a$.

But now $|\vec c| = |(-2\vec a)| = 2$, contradicting the fact that $\vec c$ is also a unit vector.

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    $\begingroup$ you could have $\vec b=-\vec a$ too, but now $\vec c$ is null vector. $\endgroup$
    – zwim
    Jan 31, 2021 at 14:43

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