Let → a and → b be the vectors along the diagonal of a parallelogram having area 2 2 . Let the angle between → a and → b be acute. | → a | = 1 and | → a . → b | =| → a × → b |. If → c = 2 2 ( → a × → b ) - 2 → b , then

Correct answer: D. (3π)/(4) Area =(1)/(2)| a × b |=2 2 | a × b |=4 2 | a |=1 and | a · b |=| a × b | θ= θ θ=(π)/(4) | a × b |=4 2 | a || b | (π)/(4)=4 2 | b |=8 Now, c =2 2 ( a × b )-2 b | c |= (2 2 )^2| a × b |^2+(2 b )^2 =16 2 Now, b · c =-2| b |^2 8 × 16 2 × α=-2.64 α=- 1 2 α=(3 π)/(4)

JEE Main 202227 JuneEvening Shift

Exam: JEE Main
Year: 2022
Date: 27 June
Shift: Evening Shift
Subject: Mathematics
Chapter: Vector Algebra

JEE Main 2022 · 27 June · Evening Shift · Mathematics · Vector Algebra

Let → a and → b be the vectors along the diagonal of a...

Let

\overset{\to}{a}

and

\overset{\to}{b}

be the vectors along the diagonal of a parallelogram having area

2\sqrt{2}

. Let the angle between

\overset{\to}{a} and \overset{\to}{b}

be acute. \left|\overset{\to}{a}\right| = 1

and \left|\overset{\to}{a} . \overset{\to}{b}\right| =\left|\overset{\to}{a} \times \overset{\to}{b}\right|. If

\overset{\to}{c} = 2\sqrt{2}\left(\overset{\to}{a} \times \overset{\to}{b}\right)

- 2\overset{\to}{b}

, then an angle between

\overset{\to}{b}

and

\overset{\to}{c}

Options

A $\dfrac{- \pi}{4}$
B $\dfrac{5\pi}{6}$
C $\dfrac{\pi}{3}$
D $\dfrac{3\pi}{4}$

Correct answer

D. $\dfrac{3\pi}{4}$

Step-by-step solution

\text{ Area }=\frac{1}{2}|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=2 \sqrt{2} \Rightarrow|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=4 \sqrt{2} |\overrightarrow{\mathrm{a}}|=1 \text{ and }|\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}|=|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}| \Rightarrow \cos \theta=\sin \theta \Rightarrow \theta=\frac{\pi}{4} \therefore|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=4 \sqrt{2} \Rightarrow|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}| \sin \frac{\pi}{4}=4 \sqrt{2} \Rightarrow|\overrightarrow{\mathrm{b}}|=8 \text{ Now, } \overrightarrow{\mathrm{c}}=2 \sqrt{2}(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})-2 \overrightarrow{\mathrm{b}} |\overrightarrow{\mathrm{c}}|=\sqrt{(2 \sqrt{2})^2|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|^2+(2 \mid \overrightarrow{\mathrm{b}})^2}=16 \sqrt{2} \text{ Now, } \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=-2|\overrightarrow{\mathrm{b}}|^2 \Rightarrow 8 \times 16 \sqrt{2} \times \cos \alpha=-2.64 \Rightarrow \cos \alpha=-\frac{1}{\sqrt{2}} \Rightarrow \alpha=\frac{3 \pi}{4}

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