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 $\overrightarrow{u}$ and $\overrightarrow{v}$ be the projection vectors of $\overrightarrow{w}=\hat{i}+\hat{j}$ along $\overrightarrow{PQ}$ and $\overrightarrow{PS}$, respectively. If $|\overrightarrow{u}|+|\overrightarrow{v}|=|\overrightarrow{w}|$ and if the area of the parallelogram $PQRS$ is 8, then which of the following statements is/are TRUE?

(A) $a+b=4$; (C) The length of the diagonal $PR$ of the parallelogram $PQRS$ is 4

(D) $\overrightarrow{w}$ is an angle bisector of the vectors $\overrightarrow{PQ}$ and $\overrightarrow{PS}$

JEE Advanced · 2020 · 27 Sep 2020 · Paper 2 · Vector Algebra

Let

a

and

b

be positive real numbers. Suppose

P Q = a \hat{i} + b \hat{j}

and

P S = a \hat{i} - b \hat{j}

are adjacent sides of a parallelogram

P QR S

. Let

u

and

v

be the projection vectors of

w = \hat{i} + \hat{j}

along

P Q

and

P S

, respectively. If

∣ u ∣ + ∣ v ∣ = ∣ w ∣

and if the area of the parallelogram

P QR S

is 8, then which of the following statements is/are TRUE?

Answer: (A), (C)

The projection magnitudes are

∣ u ∣ = \frac{a + b}{a ^{2} + b ^{2}}

and

∣ v ∣ = \frac{∣ a - b ∣}{a ^{2} + b ^{2}}

. Given

∣ u ∣ + ∣ v ∣ = ∣ w ∣ = 2

, we have

\frac{∣ a + b ∣ + ∣ a - b ∣}{a ^{2} + b ^{2}} = 2

. For

a > b

, this gives

2 a = 2 a^{2} + b^{2}

, leading to

a^{2} = b^{2}

, so

a = b

(since both are positive). The area of the parallelogram is

∣ P Q \times P S ∣ = 2 ab = 8

, so

ab = 4

. With

a = b

, we get

a = b = 2

, thus

a + b = 4

(option A). The diagonal

P R = P Q + P S = 2 a \hat{i}

, so

∣ P R ∣ = 4

(option C).

JEE Advanced · 2020 · 27 Sep 2020 · Paper 2 · One or more correct options