Question

Let $a$ and $b$ be positive real numbers. Suppose $\overrightarrow{P Q}=a \hat{i}+b \hat{j}$ and $\overrightarrow{P S}=a \hat{i}-b \hat{j}$ 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?

• A. $a+b=4$
• B. $a-b=2$
• C. The length of the diagonal $PR$ of the parallelogram $PQRS$ is $4$
• D. $\vec{ w }$ is an angle bisector of the vectors $\overrightarrow{ PQ }$ and $\overrightarrow{ PS }$

Answer 1

Answer: (A), (C)

$\vec{ u }=\frac{( a \hat{ i }+ b \hat{ j }) \cdot(\hat{ i }+\hat{ j }) \widehat{ PQ }}{\sqrt{ a ^{2}+ b ^{2}}}, \vec{ v }=\frac{( ai - b \hat{ j }) \cdot(\hat{ i }+\hat{ j }) \widehat{ PS }}{\sqrt{ a ^{2}+ b ^{2}}}$
Given $|\vec{ u }|+|\vec{ v }|=|\vec{ w }|$
$\Rightarrow |a+b|+|a-b|=\sqrt{2\left(a^{2}+b^{2}\right)} $
$=(2 a)^{2}=2\left(a^{2}+b^{2}\right)$
$ a^{2}=b^{2}$
$a \ge b$ or $a \le b$
Area $a ^{2}|(\hat{ i }+\hat{ j }) \times(\hat{ i }-\hat{ j })|=8, a =2$
Diagonal $4 i$ or $4 j$