Q75 / 150 · MHT CET 2026 (19 April Shift 1) · Mathematics
If $A(\vec{a})$, $B(\vec{b})$ and $C(\vec{c})$ are vertices of $\triangle ABC$. Point D divides segment BC internally in the ratio $2 : 1$. Point E divides segment AD internally in the ratio $1 : 2$, then the position vector of E is ____
- A$\dfrac{3\vec{a} + 4\vec{b} + 2\vec{c}}{9}$
- B$\dfrac{6\vec{a} + 2\vec{b} + \vec{c}}{9}$
- C$\dfrac{3\vec{a} + 2\vec{b} + 4\vec{c}}{9}$
- Correct answerD$\dfrac{6\vec{a} + \vec{b} + 2\vec{c}}{9}$
Solution
Point D divides the segment BC internally in the ratio $2 : 1$. Using the section formula, the position vector of D is given by:
$\vec{d} = \dfrac{2\vec{c} + 1\vec{b}}{2 + 1} = \dfrac{\vec{b} + 2\vec{c}}{3}$
Point E divides the segment AD internally in the ratio $1 : 2$. The position vector of E is given by:
$\vec{e} = \dfrac{1\vec{d} + 2\vec{a}}{1 + 2} = \dfrac{2\vec{a} + \vec{d}}{3}$
Substituting the value of $\vec{d}$ into the equation for $\vec{e}$:
$\vec{e} = \dfrac{2\vec{a} + \dfrac{\vec{b} + 2\vec{c}}{3}}{3}$
$\vec{e} = \dfrac{\dfrac{6\vec{a} + \vec{b} + 2\vec{c}}{3}}{3}$
$\vec{e} = \dfrac{6\vec{a} + \vec{b} + 2\vec{c}}{9}$
Answer: $\dfrac{6\vec{a} + \vec{b} + 2\vec{c}}{9}$
$\vec{d} = \dfrac{2\vec{c} + 1\vec{b}}{2 + 1} = \dfrac{\vec{b} + 2\vec{c}}{3}$
Point E divides the segment AD internally in the ratio $1 : 2$. The position vector of E is given by:
$\vec{e} = \dfrac{1\vec{d} + 2\vec{a}}{1 + 2} = \dfrac{2\vec{a} + \vec{d}}{3}$
Substituting the value of $\vec{d}$ into the equation for $\vec{e}$:
$\vec{e} = \dfrac{2\vec{a} + \dfrac{\vec{b} + 2\vec{c}}{3}}{3}$
$\vec{e} = \dfrac{\dfrac{6\vec{a} + \vec{b} + 2\vec{c}}{3}}{3}$
$\vec{e} = \dfrac{6\vec{a} + \vec{b} + 2\vec{c}}{9}$
Answer: $\dfrac{6\vec{a} + \vec{b} + 2\vec{c}}{9}$
Hence, the correct answer is option: 'D'.