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(D) The projection vector of $\vec{w}$ along $\vec{PQ}$ is $\vec{u} = \left( \frac{\vec{w} \cdot \vec{PQ}}{|\vec{PQ}|^2} \right) \vec{PQ}$.The magnitu

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$A, C$

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(D) The projection vector of $\vec{w}$ along $\vec{PQ}$ is $\vec{u} = \left( \frac{\vec{w} \cdot \vec{PQ}}{|\vec{PQ}|^2} ight) \vec{PQ}$.The magnitude is $|\vec{u}| = \frac{|\vec{w} \cdot \vec{PQ}|}{|\vec{PQ}|} = \frac{|(i+j) \cdot (ai+bj)|}{\sqrt{a^2+b^2}} = \frac{a+b}{\sqrt{a^2+b^2}}$.Similarly,the magnitude of the projection vector $\vec{v}$ along $\vec{PS}$ is $|\vec{v}| = \frac{|\vec{w} \cdot \vec{PS}|}{|\vec{PS}|} = \frac{|(i+j) \cdot (ai-bj)|}{\sqrt{a^2+b^2}} = \frac{|a-b|}{\sqrt{a^2+b^2}}$.Given $|\vec{u}| + |\vec{v}| = |\vec{w}| = \sqrt{1^2+1^2} = \sqrt{2}$,we have $\frac{a+b + |a-b|}{\sqrt{a^2+b^2}} = \sqrt{2}$.If $a \ge b$,then $\frac{2a}{\sqrt{a^2+b^2}} = \sqrt{2} \implies 4a^2 = 2(a^2+b^2) \implies 2a^2 = 2b^2 \implies a=b$.If $b > a$,then $\frac{2b}{\sqrt{a^2+b^2}} = \sqrt{2} \implies a=b$.Thus,$a=b$. The area of the parallelogram is $|\vec{PQ} 	imes \vec{PS}| = |(ai+bj) 	imes (ai-bj)| = |(-abk - abk)| = |-2abk| = 2ab = 8$.Since $a=b$,$2a^2 = 8 \implies a^2 = 4 \implies a=2$ (as $a>0$). Thus $a=2, b=2$.$(A)$ $a+b = 2+2 = 4$ (True).$(B)$ $a-b = 2-2 = 0 
eq 2$ (False).$(C)$ The diagonal $\vec{PR} = \vec{PQ} + \vec{PS} = (ai+bj) + (ai-bj) = 2ai = 4i$. Length is $|4i| = 4$ (True).$(D)$ $\vec{PQ} = 2i+2j$ and $\vec{PS} = 2i-2j$. The angle bisector direction is $\frac{\vec{PQ}}{|\vec{PQ}|} + \frac{\vec{PS}}{|\vec{PS}|} = \frac{2i+2j}{2\sqrt{2}} + \frac{2i-2j}{2\sqrt{2}} = \frac{4i}{2\sqrt{2}} = \sqrt{2}i$. This is along the $x$-axis,while $\vec{w} = i+j$ is at $45^\circ$. (False).

$A$. Unit vector in the direction opposite to that $a-b$ is	$(i) \ 5 \hat{i} + 3 \hat{j} - 3 \hat{k}$
$B$. If $\vec{AB} = a, \vec{BC} = b$, then $\vec{CA} =$	$(ii) \ 2 \hat{i} - \frac{8}{3} \hat{k}$
$C$. If $a, b, c$ are the position vectors of the vertices of a triangle then, its centroid is	$(iii) \ -3 \hat{i} + 4 \hat{k}$
$D$. If $d$ is a vector of magnitude $2 \sqrt{14}$ and parallel to the vector $a$, then $b + d =$	$(iv) \ -\frac{\hat{i}}{\sqrt{73}} - \frac{6 \hat{j}}{\sqrt{73}} - \frac{6 \hat{k}}{\sqrt{73}}$
	$(v) \ 3 \hat{i} + 5 \hat{j} - 3 \hat{k}$

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