Question

The points with position vectors $ 60\widehat{i}+3\widehat{j} $ , $ 40\widehat{i}-8\widehat{j} $ , $ a\widehat{i}-52\widehat{j} $ are collinear if the value of a is
A. $ -40 $
B. 40
C. $ -20 $
D. 20

Answer 1

Hint: We first take the points and name them. The points are collinear which means the lines created by those points will be similar and also parallel. Using the condition of parallel lines we find the solution for $ a $ .

Complete step by step solution:
It’s given that the points with position vectors $ 60\widehat{i}+3\widehat{j} $ , $ 40\widehat{i}-8\widehat{j} $ , $ a\widehat{i}-52\widehat{j} $ are collinear.
We first assume the vectors as $ \overrightarrow{A}=60\widehat{i}+3\widehat{j} $ , $ \overrightarrow{B}=40\widehat{i}-8\widehat{j} $ , $ \overrightarrow{C}=a\widehat{i}-52\widehat{j} $ .
We now try to find the vector lines of $ \overrightarrow{AB},\overrightarrow{AC} $ which are same line as the points are collinear.
 $ \overrightarrow{AB},\overrightarrow{AC} $ can be expressed as $ \overrightarrow{AB}=\overrightarrow{B}-\overrightarrow{A},\overrightarrow{AC}=\overrightarrow{C}-\overrightarrow{A} $ .
Therefore, $ \overrightarrow{AB}=\left( 40\widehat{i}-8\widehat{j} \right)-\left( 60\widehat{i}+3\widehat{j} \right)=\left( 40-60 \right)\widehat{i}+\left( -8-3 \right)\widehat{j}=-20\widehat{i}-11\widehat{j} $ and \[\overrightarrow{AC}=\left( a\widehat{i}-52\widehat{j} \right)-\left( 60\widehat{i}+3\widehat{j} \right)=\left( a-60 \right)\widehat{i}+\left( -52-3 \right)\widehat{j}=\left( a-60 \right)\widehat{i}-55\widehat{j}\].
These lines are similar which means they can be considered as parallel.
We know that for two parallel lines $ x\widehat{i}+y\widehat{j},m\widehat{i}+n\widehat{j} $ , we can say that $ \dfrac{x}{m}=\dfrac{y}{n} $ .
Now for $ -20\widehat{i}-11\widehat{j} $ and \[\left( a-60 \right)\widehat{i}-55\widehat{j}\], we can say $ \dfrac{-20}{a-60}=\dfrac{-11}{-55} $ .
We now sue the cross-multiplication to find the value of variable $ a $ .
Doing simplification, we get
 $ \begin{align}
  & \dfrac{-20}{a-60}=\dfrac{-11}{-55}=\dfrac{1}{5} \\
 & \Rightarrow 5\times \left( -20 \right)=a-60 \\
 & \Rightarrow a-60=-100 \\
 & \Rightarrow a=-100+60=-40 \\
\end{align} $
Therefore, the value of $ a $ is $ -40 $ .
So, the correct answer is “Option A”.

Note: If $ ab+bc=ac $ then the three points are collinear. The line segments can be translated to vectors $ ab,bc,ac $ where the magnitude of the vectors is equal to the length of the respective line segments mentioned.