Let $a$ and $b$ be positive real numbers. Suppose $\overrightarrow{PQ} = a\hat{i} + b\hat{j}$ and $\overrightarrow{PS} = a\hat{i} - b\hat{j}$ are adjacent sides of a parallelogram $PQRS$. Let $\vec{u}$ and $\vec{v}$ be the projection vectors of $\vec{w} = \hat{i} + \hat{j}$ along $\overrightarrow{PQ}$ and $\overrightarrow{PS}$, respectively. If $|\vec{u}| + |\vec{v}| = |\vec{w}|$ and if the area of the parallelogram $PQRS$ is $8$, then which of the following statements is/are TRUE?

$\vec{PQ}=a\hat{i}+b\hat{j}, \quad \vec{PS}=a\hat{i}-b\hat{j}, \quad \vec{w}=\hat{i}+\hat{j}$ $\vec{u}=\text{proj}_{\vec{PQ}}\vec{w}, \quad \vec{v}=\text{proj}_{\vec{PS}}\vec{w}, \quad |\vec{u}|+|\vec{v}|=|\vec{w}|$ $\text{Area}=|\vec{PQ}\times\vec{PS}|=8$ $\text{Which are TRUE?}$ $(1)\ a+b=4, \quad (2)\ a-b=2$ $(3)\ |\vec{PR}|=4, \quad (4)\ \vec{w} \text{ bisects } \angle QPS$ $1.\ |\vec{PQ}\times\vec{PS}|=8 \Rightarrow ab$ $2.\ |\vec{u}|,\ |\vec{v}|: \text{ projections of } \vec{w}$ $3.\ a=b; \ \text{then check (1)--(4)}$ $\vec{PQ}=(a,b,0),\quad \vec{PS}=(a,-b,0)$ $|\vec{PQ}\times\vec{PS}|=2ab=8 \Rightarrow ab=4$ $|\vec{u}|=\dfrac{|\vec{w}\cdot\vec{PQ}|}{|\vec{PQ}|}=\dfrac{|a+b|}{\sqrt{a^{2}+b^{2}}}$ $|\vec{v}|=\dfrac{|\vec{w}\cdot\vec{PS}|}{|\vec{PS}|}=\dfrac{|a-b|}{\sqrt{a^{2}+b^{2}}}$ $\dfrac{a+b+|a-b|}{\sqrt{a^{2}+b^{2}}}=|\vec{w}|=\sqrt{2}$ $a+b+|a-b|=2\max(a,b)$ $\dfrac{2\max(a,b)}{\sqrt{a^{2}+b^{2}}}=\sqrt{2} \Rightarrow a^{2}+b^{2}=2\max(a,b)^{2}$ $a^{2}+b^{2}=2\max(a,b)^{2} \Rightarrow a=b$ $ab=4 \Rightarrow a=b=2$ $\vec{PQ}=(2,2,0), \ \vec{PS}=(2,-2,0), \ \vec{PR}=(4,0,0), \ \vec{w}=(1,1,0)$ $(1)\ a+b=2+2=4 \ \checkmark$ $(2)\ a-b=2-2=0\neq 2 \ \times$ $\vec{PR}=\vec{PQ}+\vec{PS}=(2a,0,0)$ $(3)\ |\vec{PR}|=2a=4 \ \checkmark$ $\text{bisector} \parallel \dfrac{\vec{PQ}}{|\vec{PQ}|}+\dfrac{\vec{PS}}{|\vec{PS}|}=(2a,0,0)\parallel \hat{i}$ $(4)\ \vec{w}=(1,1,0)\neq \hat{i} \ \times$ $\boxed{(1)\ \text{and}\ (3)}$