Search any question & find its solution
Question:
Answered & Verified by Expert
Let $\mathrm{ABC}$ be a triangle and $\mathrm{P}$ be a point inside $\mathrm{ABC}$ such that $\overrightarrow{\mathrm{PA}}+2 \overrightarrow{\mathrm{PB}}+3 \overrightarrow{\mathrm{PC}}=\overrightarrow{0}$. The ratio of the area of triangle $\mathrm{ABC}$ to that of $\mathrm{APC}$ is-
Options:
Solution:
2676 Upvotes
Verified Answer
The correct answer is:
3
$\overrightarrow{\mathrm{PA}}+2 \overrightarrow{\mathrm{PB}}+3 \overrightarrow{\mathrm{PC}}=0$
$(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{p}})+2(\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{p}})+3(\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{p}})=0$
$\overrightarrow{\mathrm{p}}=\frac{\overrightarrow{\mathrm{a}}+2 \overrightarrow{\mathrm{b}}+3 \overrightarrow{\mathrm{c}}}{6}$
$$
\begin{array}{l}
\frac{\text { Area } \Delta \mathrm{ABC}}{\text { Area } \Delta \mathrm{APC}}=\frac{\frac{1}{2}|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}|}{\frac{1}{2}|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}|} \\
\text { put } \overrightarrow{\mathrm{p}}=\frac{\overrightarrow{\mathrm{a}}+2 \overrightarrow{\mathrm{b}}+3 \overrightarrow{\mathrm{c}}}{6} \\
\text { ratio }=3
\end{array}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.