JEE Advance - Mathematics (2020 - Paper 2 Offline - No. 11)
Let a and b be positive real numbers. Suppose $$PQ = a\widehat i + b\widehat j$$ and $$PS = a\widehat i - b\widehat j$$ are adjacent sides of a parallelogram PQRS. Let u and v be the projection vectors of $$w = \widehat i + \widehat j$$ along PQ and PS, respectively. If |u| + |v| = |w| and if the area of the parallelogram PQRS is 8, then which of the following statements is/are TRUE?
a + b = 4
a $$-$$ b = 2
The length of the diagonal PR of the parallelogram PQRS is 4
w is an angle bisector of the vectors PQ and PS
Explanation
Given vectors $$PQ = a\widehat i + b\widehat j$$ and $$PS = a\widehat i - b\widehat j$$ are adjacent sides of a parallelogram PQRS, so area of parallelogram PQRS =
|PQ $$ \times $$ PS| = 2ab = 8 (given)
$$ \therefore $$ ab = 4 ......(i)
According to the question,
|u| = |projection vector of w = $$\widehat i$$ + $$\widehat j$$ along PQ|
$$ = \left| {{{(\widehat i + \widehat j)\,.\,(a\widehat i + b\widehat j)} \over {\sqrt {{a^2} + {b^2}} }}} \right| = {{|a + b|} \over {\sqrt {{a^2} + {b^2}} }}$$
and, similarly, |v| = |projection vector of $$w = \widehat i + \widehat j$$ along PS|
$$ = \left| {{{(\widehat i + \widehat j)\,.\,(a\widehat i + b\widehat j)} \over {\sqrt {{a^2} + {b^2}} }}} \right|$$
$$ = {{|a - b|} \over {\sqrt {{a^2} + {b^2}} }} \Rightarrow |u| + |v|\, = \,|w|$$
$$ \Rightarrow {{|a + b|} \over {\sqrt {{a^2} + {b^2}} }} + {{|a - b|} \over {\sqrt {{a^2} + {b^2}} }} = \sqrt 2 $$
$$ \Rightarrow |a + b| + |a - b| = \sqrt 2 \sqrt {{a^2} + {b^2}} $$
If a $$ \ge $$ b $$ \ge $$ 0, then $$2a = \sqrt 2 \sqrt {{a^2} + {b^2}} $$
$$ \Rightarrow 4{a^2} = 2{a^2} + 2{b^2} \Rightarrow {a^2} = {b^2}$$
$$ \Rightarrow a = b$$ ....(ii)
From Eqs. (i) and (ii), we get
$$a = 2 = b \Rightarrow a + b = 4$$
and the length of diagonal PR
$$ = |a\widehat i + b\widehat j + a\widehat i - b\widehat j|$$
$$ = |2a\widehat i| = 2a = 4$$
And, the angle bisector of vector PQ and PS is along the vector
$$ \pm {\lambda \over {\sqrt {{a^2} + {b^2}} }}((a\widehat i + b\widehat j) \pm (a\widehat i - b\widehat j))$$
|PQ $$ \times $$ PS| = 2ab = 8 (given)
$$ \therefore $$ ab = 4 ......(i)
According to the question,
|u| = |projection vector of w = $$\widehat i$$ + $$\widehat j$$ along PQ|
$$ = \left| {{{(\widehat i + \widehat j)\,.\,(a\widehat i + b\widehat j)} \over {\sqrt {{a^2} + {b^2}} }}} \right| = {{|a + b|} \over {\sqrt {{a^2} + {b^2}} }}$$
and, similarly, |v| = |projection vector of $$w = \widehat i + \widehat j$$ along PS|
$$ = \left| {{{(\widehat i + \widehat j)\,.\,(a\widehat i + b\widehat j)} \over {\sqrt {{a^2} + {b^2}} }}} \right|$$
$$ = {{|a - b|} \over {\sqrt {{a^2} + {b^2}} }} \Rightarrow |u| + |v|\, = \,|w|$$
$$ \Rightarrow {{|a + b|} \over {\sqrt {{a^2} + {b^2}} }} + {{|a - b|} \over {\sqrt {{a^2} + {b^2}} }} = \sqrt 2 $$
$$ \Rightarrow |a + b| + |a - b| = \sqrt 2 \sqrt {{a^2} + {b^2}} $$
If a $$ \ge $$ b $$ \ge $$ 0, then $$2a = \sqrt 2 \sqrt {{a^2} + {b^2}} $$
$$ \Rightarrow 4{a^2} = 2{a^2} + 2{b^2} \Rightarrow {a^2} = {b^2}$$
$$ \Rightarrow a = b$$ ....(ii)
From Eqs. (i) and (ii), we get
$$a = 2 = b \Rightarrow a + b = 4$$
and the length of diagonal PR
$$ = |a\widehat i + b\widehat j + a\widehat i - b\widehat j|$$
$$ = |2a\widehat i| = 2a = 4$$
And, the angle bisector of vector PQ and PS is along the vector
$$ \pm {\lambda \over {\sqrt {{a^2} + {b^2}} }}((a\widehat i + b\widehat j) \pm (a\widehat i - b\widehat j))$$
Comments (0)
