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Acceleration in 2D

Show that the...

Question

Show that the vectors $\to a, \to b, \to c$ are co-planar if and only if $[\to a + \to b, \to b + \to c, \to c + \to a]$ are co-planar.

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Solution

$\to a, \to b, \to c$ be coplanar then $[\to a \to b \to c] = 0$
To show that $\to a + \to b, \to b + \to c, \to c + \to a$ are aslo coplanar,

consider $[\to a + \to b$ $\to b + \to c$ $\to c + \to a]$
$= (\to a + \to b) \times (\to b + \to c) . (\to c + \to a)$
$= (\to a \times \to b + \to a \times \to c + \to b \times \to b + \to b \times \to c) . (\to c + \to a)$
$= (\to a \times \to b + \to a \times \to c + \to a + \to b \to c) . (\to c + \to a)$
$[\to a \to b \to c] + [\to a \to c \to c] + [\to b \to c \to c] + [\to a \to b \to a] + [\to a \to c \to c] + [\to b \to c \to a]$
$= [\to a \to b \to c] + 0 + 0 + 0 + 0 + [\to a \to b \to c]$
$= 2 [\to a \to b \to c]$

since $[\to a + \to b$ $\to b + \to c$ $\to c + \to a] = 0$ , it follows that they are coplanar
conversely if $\to a + \to b, \to b + \to c, \to c + \to a$ are coplanar then $[\to a + \to b$ $\to b + \to c$ $\to c + \to a] = 0$

From the above it can be seen that
$[\to a + \to b$ $\to b + \to c$ $\to c + \to a] = 2 [\to a$ $\to b$ $\to c$ ]
It follows that $[\to a \to b \to c] = 0$
i.e $\to a, \to b, \to c$ are coplanar.

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