Step 1: Understanding the Question:
We need to find the ratio in which the intersection point of two cevians divides one of the cevians using section formulas or vector methods. Step 2: Key Formula or Approach:
Let $C$ be the origin $(0,0,0)$. Let $\vec{a}$ and $\vec{b}$ be position vectors of $A$ and $B$.
$\vec{P} = \frac{3\vec{c} + 4\vec{a}}{3+4}$? No, $P$ is on $AC$ such that $AP:PC = 3:4 \implies \vec{p} = \frac{4}{7}\vec{a}$.
Similarly, $BQ:QC = 3:4$ No, $CQ:QB = 4:3$ means $\vec{q} = \frac{4}{7}\vec{b}$. Step 3: Detailed Explanation:
Let $\vec{p} = \frac{3}{7}\vec{a}$ (taking $C$ as origin and $P$ on $AC$ such that $CP/CA = 3/7$).
Let $\vec{q} = \frac{4}{7}\vec{b}$ (taking $Q$ on $CB$ such that $CQ/CB = 4/7$).
Point $M$ lies on $AQ$, so $\vec{m} = (1-t)\vec{a} + t\vec{q} = (1-t)\vec{a} + \frac{4t}{7}\vec{b}$ ---- (1)
Point $M$ also lies on $BP$, so $\vec{m} = (1-s)\vec{b} + s\vec{p} = (1-s)\vec{b} + \frac{3s}{7}\vec{a}$ ---- (2)
Equating the coefficients of $\vec{a}$ and $\vec{b}$:
$1 - t = \frac{3s}{7}$ and $\frac{4t}{7} = 1 - s$.
From the second equation, $s = 1 - \frac{4t}{7}$. Substitute into the first:
$1 - t = \frac{3}{7}(1 - \frac{4t}{7}) = \frac{3}{7} - \frac{12t}{49}$
$1 - \frac{3}{7} = t - \frac{12t}{49} \implies \frac{4}{7} = \frac{37t}{49}$
$t = \frac{4}{7} \times \frac{49}{37} = \frac{28}{37}$.
The ratio in which $M$ divides $AQ$ is $t : (1-t)$:
Ratio $= \frac{28}{37} : (1 - \frac{28}{37}) = \frac{28}{37} : \frac{9}{37} = 28 : 9$. Step 4: Final Answer:
M divides AQ in the ratio $28 : 9$.
