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Question
If points A (60 \[\hat{i}\] + 3 \[\hat{j}\]), B (40 \[\hat{i}\] − 8 \[\hat{j}\]) and C (a \[\hat{i}\] − 52 \[\hat{j}\]) are collinear, then a is equal to
Options
40
−40
20
−20
Solution
−40
Given: Three points \[A\left( 60 \hat{i} + 3 \hat{j} \right), B\left( 40 \hat{i} - 8 \hat{j} \right)\] and \[C\left( a \hat{i} - 52 \hat{j} \right)\] are collinear.
Then, \[\overrightarrow{AB} = \lambda \overrightarrow{BC} .\]
We have,
\[\overrightarrow{AB} = \left( 40 \hat{i} - 8 \hat{j} \right) - \left( 60 \hat{i} + 3 \hat{j} \right) = - 20 \hat{i} - 11 \hat{j}\]
\[\overrightarrow{BC} = \left( a \hat{i} - 52 \hat{j} \right) - \left( 40 \hat{i} - 8 \hat{j} \right) = \left( a - 40 \right) \hat{i} - 44 \hat{j}\]
\[\overrightarrow{AB} = \lambda \overrightarrow{BC} \]
\[ \Rightarrow - 20 \hat{i} - 11 \hat{j} = \lambda \left( a - 40 \right) \hat{i} - \lambda44 \hat{j} \]
\[ \Rightarrow \lambda \left( a - 40 \right) = - 20 , - 44\lambda = - 11 \Rightarrow \lambda = \frac{1}{4}\]
\[ \Rightarrow a - 40 = - 80\]
\[ \Rightarrow a = - 40\]
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