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Step-by-Step Solution
Step 1: Define the resultant vector
Let
$
\vec{r} = \vec{a} + \vec{b} + \vec{c}.
$
We want to prove that
$
\vec{r} = \vec{0}.
$
Step 2: Take cross product with each vector
First, consider
$
\vec{a} \times \vec{r}.
$
We have:
$
\vec{a} \times \left(\vec{a} + \vec{b} + \vec{c}\right) = \vec{a} \times \vec{a} + \vec{a} \times \vec{b} + \vec{a} \times \vec{c}.
$
Note that
$
\vec{a} \times \vec{a} = \vec{0}.
$
Hence,
$
\vec{a} \times \vec{r} = \vec{a} \times \vec{b} + \vec{a} \times \vec{c}.
$
Step 3: Use the given condition for the cross products
It is given that
$
\vec{a} \times \vec{b} = \vec{b} \times \vec{c} = \vec{c} \times \vec{a}.
$
Rewriting
$
\vec{a} \times \vec{c}
$
as
$
-\,(\vec{c} \times \vec{a})
$
(because
$
\vec{x} \times \vec{y} = -(\vec{y} \times \vec{x})
$),
we get
$
\vec{a} \times \vec{b} + \vec{a} \times \vec{c} = \vec{a} \times \vec{b} - \vec{c} \times \vec{a}.
$
But from the given condition,
$
\vec{a} \times \vec{b} = \vec{c} \times \vec{a},
$
so they cancel each other out. Hence,
$
\vec{a} \times \vec{r} = \vec{0}.
$
Step 4: Repeat for the other vectors
Similarly, we can show:
$
\vec{b} \times \vec{r} = \vec{0}
$
and
$
\vec{c} \times \vec{r} = \vec{0}.
$
Thus,
$
\vec{r}
$
is a common vector that has zero cross product with each of
$
\vec{a}, \vec{b},
$
and
$
\vec{c}.
$
Step 5: Conclude that
$
\vec{r} = \vec{0}
$
A nonzero vector
$
\vec{r}
$
cannot have zero cross product with three non-coplanar vectors unless
$
\vec{r} = \vec{0}.
$
Hence,
$
\vec{a} + \vec{b} + \vec{c} = \vec{0}.
$
Final Answer
$
\vec{a} + \vec{b} + \vec{c} = \vec{0}.
$
