Test Index

Vector Algebra Part 2

Section: Mathematics

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Question : 88 of 100

Marks: +1, -0

Let PQR be a triangle. The points

A, B

C

are on the sides

QR, RP

and PQ respectively such that

\;\frac{QA}{AR}=\;\frac{RB}{BP}=\;\frac{PC}{CQ}=\;\frac{1}{2}

\;\frac{\;\text{Area}\,(\triangle PQR)}{\;\text{Area}\,(\triangle ABC)}

[24-Jan-2023 Shift 1]

4
3
2
$-$

Solution:

Let

P

is

\vec{0}

, Q is

\vec{q}

and

R

is

\vec{r}

A

is

\;\frac{\vec{2q + \vec{r}}}{3}, B

is

\;\frac{\vec{2r}}{3}

and

C

is

\;\frac{\vec{q}}{3}

Area of

\triangle PQR

is

= \frac{1}{2} \left| \vec{q} \times \vec{r} \right|

Area of

\triangle ABC

is

\frac{1}{2} \left| \vec{AB} \times \vec{AC} \right|

\vec{AB} = \vec{ r - \frac{2\vec{q}}{3} }, \quad \vec{AC} = \vec{ -\left( r - \frac{\vec{q}}{3} \right) }

Area of

\triangle ABC = \frac{1}{6} \left| \vec{q} \times \vec{r} \right|

\;\frac{\;\text{Area}\,(\triangle PQR)}{\text{Area}(\triangle ABC)} = 3

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