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 $|\vec w| = |\vec u| + |\vec v|$and if the area of the parallelogram PQRS is 8, then which of the following statements is/are TRUE?
A. $a + b = 4$
B. $a - b = 4$
C. Length of diagonal of parallelogram is 4
D. $\vec w$is angle bisector of $\overrightarrow {PQ} $and $\overrightarrow {PS} $

Hint: There are differences between the parallelogram's area and its vectors. Additionally, the area formulae for each are unique. Given that the data is in vector form, the area is found by first producing a matrix form for the vector, combining the two vectors into one vector form, and using this single vector.

Formula Used:
The vector cross product of two Cartesian vectors is given by, $\vec A \times \vec B = \left( {\begin{array}{*{20}{c}}
  i&j&k \\
  a&b&c \\
  e&f&g
\end{array}} \right)$
Where $\hat i,\hat j,\hat k$are unit vectors along $X.Y.Z$and $a,b,c$are components of $\vec A$and $e,f,g$are components of $\vec B$

Complete step by step Solution:
Given that,
The one side of the vector is, $\overrightarrow {PQ} = a\hat i + b\hat j$
 The other side of the vector is, $\overrightarrow {PS} = a\hat i - b\hat j$
Now, by creating the given vector in the matrix form by,
$(a\hat i - b\hat j) \times (a\hat i + b\hat j)$
$\left( {\begin{array}{*{20}{c}}
  i&j&k \\
  a&b&0 \\
  a&{ - b}&0
\end{array}} \right) = 8$
Solving we get $ab = 4$
We must take the dot product into account when determining the projection of one vector over another. A vector can be multiplied by another vector in two different ways: dot product and cross product. By resolving a vector, one can identify the components of the vector.
Therefore,
$ \Rightarrow |\vec u| = |\dfrac{{\vec w.\overrightarrow {PQ} }}{{|PQ|}}| = |\dfrac{{(\hat i + \hat j).(a\hat i + b\hat j)}}{{|a\hat i + b\hat j|}}|$
$\therefore |\vec u| = \dfrac{{(a + b)}}{{\sqrt {{a^2} + {b^2}} }}$
Similarly, $|\vec v| = |\dfrac{{\vec w.\overrightarrow {PS} }}{{|PS|}}| = |\dfrac{{(\hat i + \hat j).(a\hat i - b\hat j)}}{{|a\hat i - b\hat j|}}|$
$\therefore |\vec v| = \dfrac{{(a - b)}}{{\sqrt {{a^2} + {b^2}} }}$
Now given that $|\vec w| = |\vec u| + |\vec v|$This means
$\dfrac{{(a + b) + (a - b)}}{{\sqrt {{a^2} + {b^2}} }} = \sqrt {{1^2} + {1^2}} $
$ \Rightarrow 2a = \sqrt 2 \sqrt {{a^2} + {b^2}} $
Squaring on both sides we get,
$4{a^2} = 2{a^2} + 2{b^2}$
$\therefore a = b$ and also $ab = 4$
Therefore $a = 2 = b$
Since the diagonals from the provided parallelogram are given in the manner of $|\overrightarrow {PQ} + \overrightarrow {PS|} = |(a\hat i + b\hat j) + (a\hat i - b\hat j)|$
$\overrightarrow {|PQ} + \overrightarrow {PS} | = |2a\hat i|$
Hence length of diagonal of given parallelogram is 4

Hence, the correct option is A and C.

Note:The condition of the matrix multiplication in one matrix must be known in order to solve this problem. During the multiplications, a number of requirements must be met. The two or more vector equations are combined into a single vector equation using that matrix multiplication. Also, keep in mind that the projection is a scalar quantity which is supposed to be the length. Care should be taken while taking dot and cross product because in the projection formula we use a dot, not a cross product.