Given that vectors $a$, $b$, and $c$ are coplanar, show that the vectors $a+b$, $b+c$, and $c+a$ are also coplanar.

1. Problem Description

Given that vectors $a$, $b$, and $c$ are coplanar, show that the vectors $a+b$, $b+c$, and $c+a$ are also coplanar.

2. Solution Steps

Since the vectors $a$, $b$, and $c$ are coplanar, there exist scalars $x$ and $y$ such that

$c = xa + yb$.

We want to show that the vectors $a+b$, $b+c$, and $c+a$ are coplanar. This means that there exist scalars $\alpha$ and $\beta$ such that

$c+a = \alpha(a+b) + \beta(b+c)$.

Substituting $c = xa + yb$ into the equation above, we have:

$xa + yb + a = \alpha(a+b) + \beta(b + xa + yb)$

$xa + yb + a = \alpha a + \alpha b + \beta b + \beta xa + \beta yb$

$(x+1)a + yb = (\alpha + \beta x)a + (\alpha + \beta + \beta y)b$

Equating the coefficients of $a$ and $b$, we have the following system of equations:

$x+1 = \alpha + \beta x$

$y = \alpha + \beta + \beta y$

From the first equation, $\alpha = x+1 - \beta x$. Substituting this into the second equation:

$y = x+1 - \beta x + \beta + \beta y$

$y = x+1 + \beta(1+y-x)$

$\beta = \frac{y-x-1}{1+y-x}$

If $y-x \ne -1$, we can find a value for $\beta$, and subsequently a value for $\alpha$.

Then, $c+a = \alpha(a+b) + \beta(b+c)$, which implies that $a+b$, $b+c$, and $c+a$ are coplanar.

However, there is another approach. We can use the scalar triple product. Vectors $u, v, w$ are coplanar if and only if their scalar triple product $u \cdot (v \times w) = 0$.

Let $u = a+b$, $v = b+c$, and $w = c+a$.

Then $u \cdot (v \times w) = (a+b) \cdot ((b+c) \times (c+a))$

$= (a+b) \cdot (b \times c + b \times a + c \times c + c \times a)$

Since $c \times c = 0$, we have

$= (a+b) \cdot (b \times c + b \times a + c \times a)$

$= a \cdot (b \times c) + a \cdot (b \times a) + a \cdot (c \times a) + b \cdot (b \times c) + b \cdot (b \times a) + b \cdot (c \times a)$

We know that $a \cdot (b \times a) = 0$, $a \cdot (c \times a) = 0$, $b \cdot (b \times c) = 0$, and $b \cdot (b \times a) = 0$.

So we are left with

$= a \cdot (b \times c) + b \cdot (c \times a)$

Since $b \cdot (c \times a) = a \cdot (b \times c)$, we have

$= a \cdot (b \times c) + a \cdot (b \times c) = 2 a \cdot (b \times c)$

Since $a$, $b$, and $c$ are coplanar, $a \cdot (b \times c) = 0$.

Therefore, $u \cdot (v \times w) = 2(0) = 0$.

Thus, $a+b$, $b+c$, and $c+a$ are coplanar.

3. Final Answer

$a+b$, $b+c$, and $c+a$ are coplanar.