Question

Let $P$ be an interior point of $\Delta ABC$ such that $4\overrightarrow{PA}+3\overrightarrow{PB}+5\overrightarrow{PC}=0$. If $\text{Area}\left(\Delta ABC\right)=k\times \text{Area}\left(\Delta APC\right)$, then $k$ is

Answer 1

Let $\overrightarrow{a}=\overrightarrow{PA},\overrightarrow{b}=\overrightarrow{PB},\overrightarrow{c}=\overrightarrow{PC}$
$4\overrightarrow{a}+3\overrightarrow{b}+5\overrightarrow{c}=0\cdots \left(1\right)$

$\text{Area}\left(\Delta APC\right)=\frac{1}{2}|\overrightarrow{c}\times \overrightarrow{a}|$
$\text{Area}\left(\Delta ABC\right)=\text{Area}\left(\Delta APB\right)+\text{Area}\left(\Delta BPC\right)+\text{Area}\left(\Delta APC\right)$
$\Rightarrow \text{Area}\left(\Delta ABC\right)=\frac{1}{2}|\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{b}\times \overrightarrow{c}+\overrightarrow{c}\times \overrightarrow{a}|$

Taking cross product of equation $\left(1\right)$ with $\overrightarrow{a}$ and $\overrightarrow{c}$ one at a time, we get
$3\overrightarrow{a}\times \overrightarrow{b}=5\overrightarrow{c}\times \overrightarrow{a}$
and $3\overrightarrow{b}\times \overrightarrow{c}=4\overrightarrow{c}\times \overrightarrow{a}$
$\Rightarrow \overrightarrow{a}\times \overrightarrow{b}=\frac{5}{3}(\overrightarrow{c}\times \overrightarrow{a})$
and $\overrightarrow{b}\times \overrightarrow{c}=\frac{4}{3}(\overrightarrow{c}\times \overrightarrow{a})$

$\therefore \text{Area}\left(\Delta ABC\right)=\frac{4}{2}|\overrightarrow{c}\times \overrightarrow{a}|=4\times \text{Area}\left(\Delta APC\right)$
Hence, $k=4$