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Equality of Matrices

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Question

Show that $\to a, \to b, \to c$ are coplanar if and only if $(\to a + \to b), (\to b + \to c)$ and $(\to c + \to a)$ are coplanar.

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Solution

$\begin{matrix} \to a, \to b, \to c a r e c o p l a n a r . \Leftrightarrow [\to a \to b \to c] = 0 \Leftrightarrow 2 [\to a \to b \to c] = 0...... (1) L H S = [\to a + \to b \to b + \to c \to c + \to a] = (\to a + \to b) \cdot [(\to b + \to c) \times (\to c + \to a)] = (\to a + \to b) \cdot [\to b \times \to c + \to b \times \to a + \to c \times \to c + \to c \times \to a] = (\to a + \to b) \cdot [(\to b \times \to c) - (\to a \times \to b) + (\to c \times \to a)] {a s \to c \times \to c = 0 a n d \to b \times \to a = - \to a \times \to b} = \to a \cdot (\to b \times \to c) - \to a \cdot (\to a \times \to b) + \to a \cdot (\to c \times \to a) + \to b \cdot (\to b \times \to c) - \to b \cdot (\to a \times \to b) + \to b \cdot (\to c \times \to a) = [\to a \to b \to c] - [\to a \to a \to b] + [\to a \to c \to a] + [\to b \to b \to c] - [\to b \to a \to b] + [\to b \to c \to a] = [\to a \to b \to c] + [\to b \to c \to a] = [\to a \to b \to c] + [\to a \to b \to c] {a s [\to b \to c \to a] = [\to a \to b \to c]} = 2 [\to a \to b \to c] = R H S H e n c e, [\to a + \to b \to b + \to c \to c + \to a] = 2 [\to a \to b \to c] s o, f r o m (1), w e g e t [\to a + \to b \to b + \to c \to c + \to a] = 0 \Rightarrow \to a + \to b, \to b + \to c, \to c + \to a a r e c o p l a n a r \end{matrix}$

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