Question

for 3 points A(a), B(b), C(c) if 3a+ 2b-5c =0 then

Answer 1

C lies on the line AB and divides it in the ratio 2:3.

Answer 2

Given the equation

$$3\mathbf{a} + 2\mathbf{b} - 5\mathbf{c} = 0,$$

rearrange to express $\mathbf{c}$:

$$5\mathbf{c} = 3\mathbf{a} + 2\mathbf{b} \quad \Longrightarrow \quad \mathbf{c} = \frac{3\mathbf{a} + 2\mathbf{b}}{5}.$$

This shows that the position vector of $C$ is a weighted average of the position vectors of $A$ and $B$. According to the section formula, if a point $C$ divides the line segment $AB$ internally in the ratio $AC:CB = m:n$, then

$$\mathbf{c} = \frac{n\mathbf{a} + m\mathbf{b}}{m+n}.$$

Comparing with $\mathbf{c} = \frac{3\mathbf{a} + 2\mathbf{b}}{5}$, we have $n = 3$ and $m = 2$. Therefore, the ratio $AC:CB = 2:3$.

Thus, the points $A$, $B$, and $C$ are collinear, with $C$ dividing $AB$ in the ratio $2:3$.