A, B and C are three distinct points with position vectors \(\vec{a}, \vec{b}\) and \(\vec{c}\) respectively, such that \(3\vec{a}-5\vec{b}+2\vec{c}=\vec{0}\). Then
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → B divides AC internally in the ratio 2 : 3
The coefficients 3 and 2 have the same sign, so the relation is written with \(\vec{b}\) alone on one side:
\(5\vec{b} = 3\vec{a}+2\vec{c}\)
\( \displaystyle \vec{b} = \frac{2\vec{c}+3\vec{a}}{2+3} \)
By the section formula, \(\frac{m\vec{c}+n\vec{a}}{m+n}\) is the point dividing AC internally in the ratio m : n. Here m = 2 and n = 3, so B divides AC internally in the ratio 2 : 3.
Why the other options are incorrect
Option (1) C divides AB internally in the ratio 5 : 3 — Incorrect. From the relation, \(\vec{c} = \frac{5\vec{b}-3\vec{a}}{5-3}\), which is division of AB externally in the ratio 5 : 3.
Option (2) B divides AC internally in the ratio 3 : 2 — Incorrect. That point is \(\frac{3\vec{c}+2\vec{a}}{5}\); the weights on \(\vec{a}\) and \(\vec{c}\) are interchanged.
Option (4) A divides BC internally in the ratio 2 : 5 — Incorrect. From the relation, \(\vec{a} = \frac{5\vec{b}-2\vec{c}}{3}\), which has a negative weight, so A lies outside the segment BC.