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Question: The points with position vectors \( 60\widehat{i}+3\widehat{j} \) , \( 40\widehat{i}-8\widehat{j} \)...

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

Explanation

Solution

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 aa .

Complete step by step solution:
It’s given that the points with position vectors 60i^+3j^60\widehat{i}+3\widehat{j} , 40i^−8j^40\widehat{i}-8\widehat{j} , ai^−52j^a\widehat{i}-52\widehat{j} are collinear.
We first assume the vectors as A→=60i^+3j^\overrightarrow{A}=60\widehat{i}+3\widehat{j} , B→=40i^−8j^\overrightarrow{B}=40\widehat{i}-8\widehat{j} , C→=ai^−52j^\overrightarrow{C}=a\widehat{i}-52\widehat{j} .
We now try to find the vector lines of AB→,AC→\overrightarrow{AB},\overrightarrow{AC} which are same line as the points are collinear.
AB→,AC→\overrightarrow{AB},\overrightarrow{AC} can be expressed as AB→=B→−A→,AC→=C→−A→\overrightarrow{AB}=\overrightarrow{B}-\overrightarrow{A},\overrightarrow{AC}=\overrightarrow{C}-\overrightarrow{A} .
Therefore, AB→=(40i^−8j^)−(60i^+3j^)=(40−60)i^+(−8−3)j^=−20i^−11j^\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 AC→=(ai^−52j^)−(60i^+3j^)=(a−60)i^+(−52−3)j^=(a−60)i^−55j^\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 xi^+yj^,mi^+nj^x\widehat{i}+y\widehat{j},m\widehat{i}+n\widehat{j} , we can say that xm=yn\dfrac{x}{m}=\dfrac{y}{n} .
Now for −20i^−11j^-20\widehat{i}-11\widehat{j} and (a−60)i^−55j^\left( a-60 \right)\widehat{i}-55\widehat{j}, we can say −20a−60=−11−55\dfrac{-20}{a-60}=\dfrac{-11}{-55} .
We now sue the cross-multiplication to find the value of variable aa .
Doing simplification, we get
−20a−60=−11−55=15 ⇒5×(−20)=a−60 ⇒a−60=−100 ⇒a=−100+60=−40 \begin{aligned} & \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{aligned}
Therefore, the value of aa is −40-40 .
So, the correct answer is “Option A”.

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