# Three unit vectors whose sum is zero

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

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)

• $c = -a -b$, 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. Jan 31, 2021 at 14:38

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?)

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

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.

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