The points with position vectors $60 \hat{i}+3 \hat{j}, 40 \hat{i}-8 \hat{j}, a \hat{i}-52 \hat{j}$

Question

The points with position vectors $60 \hat{i}+3 \hat{j}, 40 \hat{i}-8 \hat{j}, a \hat{i}-52 \hat{j}$ are collinear if\

Professor PC Thomas Classes · Answer

(a) : Let $A, B, C$ be the given points. Then $\overrightarrow{O A}=60 \hat{i}+3 \hat{j}$;\ $\overrightarrow{O B}=40 \hat{i}-8 \hat{j} ; \overrightarrow{O C}=a \hat{i}-52 \hat{j}$\ $\overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A}=(40 \hat{i}-8 \hat{j})-(60 \hat{i}+3 \hat{j})$\ $=-20 \hat{i}-11 \hat{j}$\ $\overrightarrow{A C}=\overrightarrow{O C}-\overrightarrow{O A}=(a \hat{i}-52 \hat{j})-(40 \hat{i}-8 \hat{j})$\ $=(a-40) \hat{i}-44 \hat{j}$\ Since $A, B, C$ are collinear, $\overrightarrow{A C}=\lambda \overrightarrow{A B}$ for some $\lambda \in \mathbb{R}$\ $\Rightarrow \quad(a-40) \hat{i}-44 \hat{j}=\lambda[-20 \hat{i}-11 \hat{j}]$\ $\Rightarrow \quad a-40=-20 \lambda$ and $-44=-11 \lambda$\ $	herefore \quad \lambda=4, a=-40$\