$\vec{a}$ and $\vec{b}$ are two non-collinear vectors then the points with position vectors $l_1 \ve

Question

$\vec{a}$ and $\vec{b}$ are two non-collinear vectors then the points with position vectors $l_1 \vec{a}+m_1 \vec{b}, l_2 \vec{a}+m_2 \vec{b}, l_3 \vec{a}+m_3 \vec{b}$ are collinear then find the value of $\left|\begin{array}{lll}1 & 1 & 1 \ l_1 & l_2 & l_3 \ m_1 & m_2 & m_3\end{array}ight|$.

Professor PC Thomas Classes · Answer

$
\vec{r}_1=l_1 \vec{a}+m_1 \vec{b} ; \vec{r}_2=l_2 \vec{a}+m_2 \vec{b} ; \vec{r}_3=l_3 \vec{a}+m_3 \vec{b}
$
Since $\vec{r}_1, \vec{r}_2$ and $\vec{r}_2$ are co-linear $\left(\vec{r}_1-\vec{r}_2ight) 	imes\left(\vec{r}_2-\vec{r}_3ight)=0$
$\left(\left(l_1-l_2ight) \vec{a}+\left(m_1-m_2ight) \vec{b}ight) 	imes\left(\left(l_2-l_3ight) \vec{a}+\left(m_2-m_3ight) \vec{b}ight)=0$
$\left(l_1-l_2ight)\left(m_2-m_3ight)(\vec{a} 	imes \vec{b})-\left(m_1-m_2ight)\left(l_2-l_3ight)(\vec{a} 	imes \vec{b})=0$ or
$\left(l_1-l_2ight)\left(m_2-m_3ight)-\left(m_1-m_2ight)\left(l_2-l_3ight)=0$
$\left|\begin{array}{ccc}1 & 1 & 1 \ l_1 & l_2 & l_3 \ m_1 & m_2 & m_3\end{array}ight|=\left|\begin{array}{ccc}1 & 1 & 1 \ l_1-l_2 & l_2 & l_3 \ m_1-m_2 & m_2 & m_3\end{array}ight|=\left|\begin{array}{ccc}0 & 1 & 1 \ l_1-l_2 & l_2-l_3 & l_3 \ m_1-m_2 & m_2-m_3 & m_3\end{array}ight|$
$\left(l_1-l_2ight)\left(m_2-m_3ight)-\left(m_1-m_2ight)\left(l_2-l_3ight)=0$