JEE Advanced 2020 — Paper 2 — Question 44

Given $\vec{PQ} = a\hat{i} + b\hat{j}$, $\vec{PS} = a\hat{i} - b\hat{j}$, and $\vec{w} = \hat{i} + \hat{j}$. Compute projection vectors: $\vec{u} = \frac{\vec{w}\cdot\vec{PQ}}{|\vec{PQ}|^2}\vec{PQ} = \frac{a+b}{a^2+b^2}(a\hat{i}+b\hat{j})$, so $|\vec{u}| = \frac{a+b}{\sqrt{a^2+b^2}}$.

Similarly $|\vec{v}| = \frac{|a-b|}{\sqrt{a^2+b^2}}$. Given $|\vec{u}|+|\vec{v}|=|\vec{w}| = \sqrt{2}$, we have $\frac{a+b+|a-b|}{\sqrt{a^2+b^2}} = \sqrt{2}$. If $a\ge b$, then numerator = $2a$, giving $2a = \sqrt{2}\sqrt{a^2+b^2}$.

Squaring yields $4a^2 = 2(a^2+b^2)$ → $a^2 = b^2$. Since $a,b >0$, $a=b$.

The case $b>a$ leads to $a=b$, contradiction. Hence $a=b$. Area of parallelogram = $|\vec{PQ}\times\vec{PS}| = |(a\hat{i}+b\hat{j})\times(a\hat{i}-b\hat{j})| = | -2ab \hat{k}| = 2ab = 2a^2 = 8$ → $a^2=4$ → $a=2$. Thus $a=b=2$. Check options:

A) $a+b=4$ (true).

B) $a-b=2$ (false).

C) Diagonal $\vec{PR} = \vec{PQ}+\vec{PS} = 4\hat{i}$, length = 4 (true).

D) $\vec{w} = \hat{i}+\hat{j}$ is parallel to $\vec{PQ}$ but not the angle bisector of $\vec{PQ}$ and $\vec{PS}$; the bisector is along $\hat{i}$. Hence D is false. Correct statements: A and C.