Let a and b be positive real numbers. Suppose → PQ = a ^ i + b ^ j and → PS = a ^ i - b ^ j are adjacent sides of a parallelogram PQRS . Let → u and → v be the projection vectors of → w = ^ i + ^ j along → PQ and → PS

Correct answer: A. a + b = 4 Projection of → a along → b = → a . → b | → b | · ^ b → u = \ ( ^ i + ^ j ) · a ^ i + b ^ j a ^ 2 + b ^ 2 \ a ^ i + b ^ j a ^ 2 + b ^ 2 = (a + b) a ^ 2 + b ^ 2 · a ^ i + b ^ j a ^ 2 + b ^ 2 → v = \ ( ^ i + ^ j ) · a ^ i - b ^ j a ^ 2 + b ^ 2 \ a ^ i - b ^ j a ^ 2 + b ^ 2 = (a - b) a ^ 2 + b ^ 2 · a ^ i - b ^ j a ^ 2 + b ^ 2 given | cases i & j & k a & b & 0 a & - b & 0 cases | = 8 | ^ k (- ab - ab)| = 8 ab = 4 Given | → u | + | → v | = → w a + b a ^ 2 + b ^ 2 |a - b| a ^ 2 + b ^ 2 = 2 For a > b , 2a = 2 a ^ 2 + b ^ 2 2 a ^ 2 = a ^ 2 + b ^ 2 a ^ 2 = b ^ 2 ⇒ a = b = 2 Similarly, for b > a, a = b = 2 a + b = 4 a - b = 0 Length of diagonal → |PR| = | → PS + → PQ | = |2a ^ i | = 2a = 4.

JEE Advanced 2020Paper 2

Exam: JEE Advanced
Year: 2020
Paper: Paper 2
Subject: Mathematics
Chapter: Vector Algebra

JEE Advanced 2020 · Paper 2 · Mathematics · Vector Algebra

Let a and b be positive real numbers. Suppose → PQ = a ^ i + b ^ j...

Let

a

and

b

be positive real numbers. Suppose

\overset{\to}{PQ} = a\overset{^}{i} + b\overset{^}{j}

and

\overset{\to}{PS} = a\overset{^}{i} - b\overset{^}{j}

are adjacent sides of a parallelogram

PQRS .

Let

\overset{\to}{u}

and

\overset{\to}{v}

be the projection vectors of

\overset{\to}{w} = \overset{^}{i} + \overset{^}{j}

along

\overset{\to}{PQ}

and

\overset{\to}{PS}

, respectively. If

| \overset{\to}{u} | + \left|\overset{\to}{v}\right| = \left|\overset{\to}{w}\right|

and if the area of the parallelogram

PQRS

8,

then which of the following statements is/are TRUE? Let a and b be positive real numbers. Suppose → PQ = a ^ i + b ^ j and → PS = a ^ i - b ^ j are adjacent sides of a

Options

A $a + b = 4$
B $a - b = 2$
CThe length of the diagonal $PR$ of the parallelogram $PQRS$ is 4
D $\overset{\to}{w}$ is an angle bisector of the vectors $\overset{\to}{PQ}$ and $\overset{\to}{PS}$

Correct answer

A. $a + b = 4$

Step-by-step solution

Let a and b be positive real numbers. Suppose → PQ = a ^ i + b ^ j and → PS = a ^ i - b ^ j are adjacent sides of a

Projection of $\overset{\to}{a}along\overset{\to}{b} = \dfrac{\overset{\to}{a} . \overset{\to}{b}}{\left|\overset{\to}{b}\right|} \cdot \overset{^}{b}$

$\overset{\to}{u} = \left\{\left(\overset{^}{i} + \overset{^}{j}\right) \cdot \dfrac{a\overset{^}{i} + b\overset{^}{j}}{\sqrt{{a}^{2} + {b}^{2}}}\right\}\dfrac{a\overset{^}{i} + b\overset{^}{j}}{\sqrt{{a}^{2} + {b}^{2}}} = \dfrac{\left(a + b\right)}{\sqrt{{a}^{2} + {b}^{2}}} \cdot \dfrac{a\overset{^}{i} + b\overset{^}{j}}{\sqrt{{a}^{2} + {b}^{2}}}$

$\overset{\to}{v} = \left\{\left(\overset{^}{i} + \overset{^}{j}\right) \cdot \dfrac{a\overset{^}{i} - b\overset{^}{j}}{\sqrt{{a}^{2} + {b}^{2}}}\right\}\dfrac{a\overset{^}{i} - b\overset{^}{j}}{\sqrt{{a}^{2} + {b}^{2}}} = \dfrac{\left(a - b\right)}{\sqrt{{a}^{2} + {b}^{2}}} \cdot \dfrac{a\overset{^}{i} - b\overset{^}{j}}{\sqrt{{a}^{2} + {b}^{2}}}$

given \left| $\begin{cases}i & j & k \\ a & b & 0 \\ a & - b & 0\end{cases}$ \right| = 8 $\left|\overset{^}{k}\left(- ab - ab\right)\right| = 8$

$ab = 4$

Given \left|\overset{\to}{u}\right| + | \overset{\to}{v} | = \overset{\to}{w} $$ \dfrac{a + b}{\sqrt{{a}^{2} + {b}^{2}}}\dfrac{\left|a - b\right|}{\sqrt{{a}^{2} + {b}^{2}}} = \sqrt{2} $For$ a > b $,$ 2a = \sqrt{2}\sqrt{{a}^{2} + {b}^{2}} $$ 2{a}^{2} = {a}^{2} + {b}^{2} $$ {a}^{2} = {b}^{2} ⇒ a = b = 2 $Similarly, for$ b > a, a = b = 2 $$ a + b = 4 $$ a - b = 0 $Length of diagonal$ \overset{\to}{\left|PR\right|} = \left|\overset{\to}{PS} + \overset{\to}{PQ}\right| = \left|2a\overset{^}{i}\right| = 2a = 4.

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