JEE MAIN - Mathematics (2023 - 24th January Morning Shift - No. 12)
Let PQR be a triangle. The points A, B and C are on the sides QR, RP and PQ respectively such that
$${{QA} \over {AR}} = {{RB} \over {BP}} = {{PC} \over {CQ}} = {1 \over 2}$$. Then $${{Area(\Delta PQR)} \over {Area(\Delta ABC)}}$$ is equal to :
$${{QA} \over {AR}} = {{RB} \over {BP}} = {{PC} \over {CQ}} = {1 \over 2}$$. Then $${{Area(\Delta PQR)} \over {Area(\Delta ABC)}}$$ is equal to :
$$\frac{5}{2}$$
4
2
3
Explanation
Let the position vector of $P, Q, R$ be $\overrightarrow{0}, \overrightarrow{a}, \overrightarrow{b}$
$\Rightarrow$ Position vector of $A=\frac{2 \overrightarrow{a}+\overrightarrow{b}}{3}, $
Position vector of $B=\frac{2 \overrightarrow{b}}{3}$ and
Position vector of $C=\frac{\overrightarrow{a}}{3}$
_24th_January_Morning_Shift_en_12_1.png)
$$ \begin{aligned} & \therefore \overrightarrow{A B}=\frac{2 \vec{b}}{3}-\left(\frac{2 \vec{a}+\vec{b}}{3}\right)=\frac{\vec{b}}{3}-\frac{2 \vec{a}}{3}=\frac{\vec{b}-2 \vec{a}}{3} \\\\ & \overrightarrow{C A}=\frac{2 \vec{a}+\vec{b}}{3}-\frac{\vec{a}}{3}=\frac{\vec{a}+\vec{b}}{3} \\\\ & \text { Area of } \triangle P Q R=\frac{1}{2}|\overrightarrow{P Q} \times \overrightarrow{P R}|=\frac{1}{2}|\vec{a} \times \vec{b}| \\\\ & \text { Area of } \triangle A B C=\frac{1}{2}|\overrightarrow{C A} \times \overrightarrow{A B}| \\ & =\frac{1}{2}\left|\left(\frac{\vec{a}+\vec{b}}{3}\right) \times\left(\frac{\vec{b}-2 \vec{a}}{3}\right)\right|=\frac{1}{2}\left|\frac{\vec{a} \times \vec{b}}{3}\right| \\\\ & \therefore \frac{\text { Area }(\triangle P Q R)}{\text { Area }(\triangle A B C)}=\frac{\frac{1}{2}|\vec{a} \times \vec{b}|}{\frac{1}{2}\left|\frac{\vec{a} \times \vec{b}}{3}\right|}=3 \text { sq. units } \end{aligned} $$
$\Rightarrow$ Position vector of $A=\frac{2 \overrightarrow{a}+\overrightarrow{b}}{3}, $
Position vector of $B=\frac{2 \overrightarrow{b}}{3}$ and
Position vector of $C=\frac{\overrightarrow{a}}{3}$
$$ \begin{aligned} & \therefore \overrightarrow{A B}=\frac{2 \vec{b}}{3}-\left(\frac{2 \vec{a}+\vec{b}}{3}\right)=\frac{\vec{b}}{3}-\frac{2 \vec{a}}{3}=\frac{\vec{b}-2 \vec{a}}{3} \\\\ & \overrightarrow{C A}=\frac{2 \vec{a}+\vec{b}}{3}-\frac{\vec{a}}{3}=\frac{\vec{a}+\vec{b}}{3} \\\\ & \text { Area of } \triangle P Q R=\frac{1}{2}|\overrightarrow{P Q} \times \overrightarrow{P R}|=\frac{1}{2}|\vec{a} \times \vec{b}| \\\\ & \text { Area of } \triangle A B C=\frac{1}{2}|\overrightarrow{C A} \times \overrightarrow{A B}| \\ & =\frac{1}{2}\left|\left(\frac{\vec{a}+\vec{b}}{3}\right) \times\left(\frac{\vec{b}-2 \vec{a}}{3}\right)\right|=\frac{1}{2}\left|\frac{\vec{a} \times \vec{b}}{3}\right| \\\\ & \therefore \frac{\text { Area }(\triangle P Q R)}{\text { Area }(\triangle A B C)}=\frac{\frac{1}{2}|\vec{a} \times \vec{b}|}{\frac{1}{2}\left|\frac{\vec{a} \times \vec{b}}{3}\right|}=3 \text { sq. units } \end{aligned} $$
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