Vectors & 3D GeometrymediumPYQ · JEE Main · 5 Apr 2026 · Shift 2 (Afternoon)Free

Vectors & 3D Geometry: Let Origin Equals (JEE Main 2026)

JEE Maths question with a full step-by-step solution.

Question
Let OO be the origin with OP→=a⃗\overrightarrow{OP}=\vec{a} and OQ→=b⃗\overrightarrow{OQ}=\vec{b}. If RR is on OP→\overrightarrow{OP} such that OP→=5OR→\overrightarrow{OP}=5\overrightarrow{OR}, and MM is such that OQ→=5RM→\overrightarrow{OQ}=5\overrightarrow{RM}, then PM→\overrightarrow{PM} equals:
A15(a⃗−4b⃗)\dfrac{1}{5}(\vec{a}-4\vec{b})
B15(b⃗−4a⃗)\dfrac{1}{5}(\vec{b}-4\vec{a})correct
C15(−a⃗+4b⃗)\dfrac{1}{5}(-\vec{a}+4\vec{b})
D15(−b⃗+4a⃗)\dfrac{1}{5}(-\vec{b}+4\vec{a})
Solution
Step 1: Express OR→\overrightarrow{OR}
OP→=5OR→  ⟹  OR→=a⃗5\overrightarrow{OP} = 5\overrightarrow{OR} \implies \overrightarrow{OR} = \frac{\vec{a}}{5}
 Step 2: Express OM→\overrightarrow{OM}
OQ→=5RM→  ⟹  RM→=b⃗5\overrightarrow{OQ} = 5\overrightarrow{RM} \implies \overrightarrow{RM} = \frac{\vec{b}}{5}
OM→=OR→+RM→=a⃗5+b⃗5=a⃗+b⃗5\overrightarrow{OM} = \overrightarrow{OR}+\overrightarrow{RM} = \frac{\vec{a}}{5}+\frac{\vec{b}}{5} = \frac{\vec{a}+\vec{b}}{5}
 Step 3: Compute PM→\overrightarrow{PM}
PM→=OM→−OP→=a⃗+b⃗5−a⃗=a⃗+b⃗−5a⃗5=15(b⃗−4a⃗)\overrightarrow{PM} = \overrightarrow{OM}-\overrightarrow{OP} = \frac{\vec{a}+\vec{b}}{5}-\vec{a} = \frac{\vec{a}+\vec{b}-5\vec{a}}{5} = \frac{1}{5}(\vec{b}-4\vec{a})
 Answer: (2)
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