Question

Let $a$ and $b$ be positive real numbers. Suppose $\overrightarrow{PQ}=a\hat{i}+b\hat{j}\text{and}\overrightarrow{PS}=a\hat{i}-b\hat{j}$​ are adjacent sides of a parallelogram $PQRS$. Let $u$ and$v$ be the projection vectors of $\overrightarrow{w}=\hat{i}+{\hat{j}}^{}$ along $\overrightarrow{PQ}\text{and}\overrightarrow{PS}$ respectively. If $\mid u\mid +\mid v\mid =\mid w\mid$ 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 $\overrightarrow{PR}$ of the parallelogram is $4$
• D. $\overrightarrow{w}$is an angle bisector of the vectors $\overrightarrow{PQ}\text{and}\overrightarrow{PS}$

Answer 1

Answer: (A), (C)

The length of the diagonal $\overrightarrow{PR}$ of the parallelogram is $4$

Explanation for the correct options:

Finding which of the given statement is true

Given: A parallelogram $PQRS$ with $\overrightarrow{PQ}=a\hat{i}+b\hat{j}\text{and}\overrightarrow{PS}=a\hat{i}-b\hat{j}$​ as adjacent sides.

$$\begin{gathered} (a\hat{i}+b{\hat{j}}^{})\times (a\hat{i}-b\widehat{j^{}})=0 \\ \left[\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ a& b& 0\\ a& -b& 0\end{array}\right]=8 \\ ab=4 \\ \left|u\right|+\left|v\right|=\left|w\right| \end{gathered}$$

$$\begin{gathered} \overrightarrow{u}\left\{w.(\overrightarrow{PQ})\right\}(\overrightarrow{PQ}) \\ \overrightarrow{v}=(w.\overrightarrow{PQ})\}\overrightarrow{PQ} \\ \overrightarrow{u}=\left(\frac{a+b}{\sqrt{a^2+b^2}}\right)\frac{a\hat{i}+b\hat{j}}{\sqrt{a^2+b^2}} \\ \Rightarrow \left(\frac{a+b}{\sqrt{a^2+b^2}}\right)a\hat{i}+b\hat{j} \\ \overrightarrow{v}=\left(\frac{a-b}{a^2+b^2}\right)a\hat{i}-b\hat{j}\left|w\right|=2 \end{gathered}$$

$$\begin{gathered} \frac{(a^2-ab)}{a^2+b^2}-\frac{(ab-b^2)}{a^2+b^2}\hat{j} \\ \Rightarrow \frac{2a}{\sqrt{a^2+b^2}}=\sqrt{2} \end{gathered}$$

$\begin{array}{rcl}& & \left|\overrightarrow{u}\right|=\frac{a+b}{a^2+b^2}\Rightarrow \frac{1}{a^2+b^2}\sqrt{({(a-b)a)}^2+({(a-b)b)}^2}\\ & & \left|\overrightarrow{v}\right|=\frac{a-b}{a^2+b^2}\Rightarrow \frac{(a-b)}{a^2+b^2}\\ \sqrt{2}\sqrt{a^2+b^2}& =& 2aab=4\\ 2a^2+2b^2& =& 4a^{2}a=b=2\\ b^2& =& a^2\\ a+b& =& 0\\ a& =& b=0\\ & \Rightarrow & \overrightarrow{PS}+\overrightarrow{PQ}=2a\hat{i}\end{array}$

Length$=2a=4$

Hence the correct answer are options (A) and (C).