The position vector of a point which divides the line joining the points with position vectors \( (\vec{a} - 2\vec{b}) \) and \( (2\vec{a} + \vec{b}) \) externally in the ratio \( 2 : 1 \), is

\( 3\vec{a} + 4\vec{b} \)

The correct answer is Option (1) → \( 3\vec{a} + 4\vec{b} \)

Given:

• Point A has position vector $\vec{A} = \vec{a} - 2\vec{b}$
• Point B has position vector $\vec{B} = 2\vec{a} + \vec{b}$
• The required point divides the line joining A and B externally in the ratio $2:1$

Formula for external division of position vectors:

If point $P$ divides $\vec{A}$ and $\vec{B}$ externally in the ratio $m:n$, then

$\vec{P} = \frac{m\vec{B} - n\vec{A}}{m - n}$

Using $m = 2$, $n = 1$:

$\vec{P} = \frac{2(2\vec{a} + \vec{b}) - 1(\vec{a} - 2\vec{b})}{2 - 1}$
$= \frac{4\vec{a} + 2\vec{b} - \vec{a} + 2\vec{b}}{1}$
$= (4\vec{a} - \vec{a}) + (2\vec{b} + 2\vec{b}) = 3\vec{a} + 4\vec{b}$