Let O be the origin, $\overrightarrow{\mathrm{OP}}=\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{OQ}}=\overrightarrow{\mathrm{b}}$. If $R$ is the point on $\overrightarrow{\mathrm{OP}}$ such that $\overrightarrow{\mathrm{OP}}=5 \overrightarrow{\mathrm{OR}}$, and M is the point such that $\overrightarrow{\mathrm{OQ}}=5 \overrightarrow{\mathrm{RM}}$, then $\overrightarrow{\mathrm{PM}}$ is equal to:

(B) $\frac{1}{5}(\vec{b}-4 \vec{a})$

(A) $\frac{1}{5}(\vec{a}-4 \vec{b})$

(C) $\frac{1}{5}(-\vec{a}+4 \vec{b})$

(D) $\frac{1}{5}(-\vec{b}+4 \vec{a})$

JEE Main · 2026 · 05 Apr 2026 · Shift 2 · Vector Algebra

Let O be the origin,

OP = a

and

OQ = b

. If

R

is the point on

OP

such that

OP = 5 OR

, and M is the point such that

OQ = 5 RM

, then

PM

is equal to:

Answer: (B)

O P = 5 O R

gives

O R = \frac{1}{5} a

, and

O Q = 5 R M

gives

R M = \frac{1}{5} b

. Hence

O M = O R + R M = \frac{1}{5} (a + b)

, and

P M = O M - O P = \frac{1}{5} (a + b) - a = \frac{1}{5} (b - 4 a)

. Thus

P M = \frac{1}{5} (b - 4 a)

JEE Main · 2026 · 05 Apr 2026 · Shift 2 · Single correct option