Exams Joint Entrance Examination Mathematics Vector Algebra Question 3 Question 3 of 379 Easy MCQ
Let O O O be the origin, O P → ← = a ⃗ \overset{\to \leftarrow }{OP}=\vec{a} O P →← = a and O Q → ← = b ⃗ \overset{\to \leftarrow }{OQ}=\vec{b} O Q →← = b . If R R R is the point on O P → ← \overset{\to \leftarrow }{OP} O P →← such that O P → ← = 5 O R → ← \overset{\to \leftarrow }{OP}=5\overset{\to \leftarrow }{OR} O P →← = 5 O R →← , and M M M is the point such that O Q → ← = 5 R M → ← \overset{\to \leftarrow }{OQ}=5\overset{\to \leftarrow }{RM} O Q →← = 5 R M →← , then P M → ← \overset{\to \leftarrow }{PM} P M →← is equal to :
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O R → ← = O P → ← 5 = a → 5 R M → ← = O Q → ← 5 = b → 5 O M → ← − O R → ← = b → 5 O M → ← = a → + b ⃗ 5 P M → ← = O M → ← − O P → ← a → + b → 5 − a → = b → − 4 a → 5 \begin{gathered} \overset{\to \leftarrow }{OR}=\frac{\overset{\to \leftarrow }{OP}}{5}=\frac{\overset{\to }{a}}{5} \\ \overset{\to \leftarrow }{RM}=\frac{\overset{\to \leftarrow }{OQ}}{5}=\frac{\overset{\to }{b}}{5} \\ \overset{\to \leftarrow }{OM}-\overset{\to \leftarrow }{OR}=\frac{\overset{\to }{b}}{5} \\ \overset{\to \leftarrow }{OM}=\frac{\overset{\to }{a}+\vec{b}}{5} \\ \overset{\to \leftarrow }{PM}=\overset{\to \leftarrow }{OM}-\overset{\to \leftarrow }{OP} \\ \frac{\overset{\to }{a}+\overset{\to }{b}}{5}-\overset{\to }{a} \\ =\frac{\overset{\to }{b}-4\overset{\to }{a}}{5} \end{gathered} O R →← = 5 O P →← = 5 a → R M →← = 5 O Q →← = 5 b → O M →← − O R →← = 5 b → O M →← = 5 a → + b P M →← = O M →← − O P →← 5 a → + b → − a → = 5 b → − 4 a → Answer the question to unlock