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Collinear Vectors

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Question

If $\to a$ and $\to b$ are two non-collinear vectors, show that points $l_{1} \to a + m_{1} \to b, l_{2} \to a + m_{2} \to b$ and $l_{3} \to a + m_{3} \to b$ are collinear, if $∣ ∣ ∣ ∣ \begin{matrix} l_{1} & l_{2} & l_{3} m_{1} & m_{2} & m_{3} 1 & 1 & 1 \end{matrix} ∣ ∣ ∣ ∣$ equals to:

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Solution

$\to a \times \to b \neq 0$
$A = l_{1} \to a + m_{1} \to b$
$B = l_{2} \to a + m_{2} \to b$
$C = l_{3} \to a + m_{3} \to b$
$A B = (l_{1} - l_{2}) \to a + (m_{1} - m_{2}) \to b$
$B C = (l_{3} - l_{2}) \to a + (m_{3} - m_{2}) \to b$
$- - \to B A \times - - \to B C = (l_{1} l_{2}) (m_{3} - m_{2}) (\to a \times \to b) - (m_{1} m_{2}) (l_{3} - l_{2}) (\to a \times \to b) + 0 = 0$
$\Rightarrow (l_{1} - l_{2}) (m_{3} - m_{2}) - (m_{1} - m_{2}) (l_{3} - l_{2}) = 0$
$\Rightarrow l_{1} m_{3} - l_{2} m_{3} - l_{1} m_{2} - m_{1} l_{3} + m_{1} l_{2} + m_{2} l_{3} = 0$
$= ∣ ∣ ∣ ∣ \begin{matrix} l_{1} & l_{2} & l_{3} m_{1} & m_{2} & m_{3} 1 & 1 & 1 \end{matrix} ∣ ∣ ∣ ∣$

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