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JEE Main 202225 JuneMorning Shift
Exam
JEE Main
Year
2022
Date
25 June
Shift
Morning Shift
Subject
Mathematics
Chapter
Vector Algebra

JEE Main 2022 · 25 June · Morning Shift · Mathematics · Vector Algebra

Let θ be the angle between the vectors → a and → b , where | → a |...

Let θ\theta be the angle between the vectors a→andb→\overset{\to}{a} and \overset{\to}{b}, where \left|\overset{\to}{a}\right| = 4, \left|\overset{\to}{b}\right| = 3andand\theta \in \left(\dfrac{\pi}{4}, \dfrac{\pi}{3}\right).Then. Then{\left|\left(\overset{\to}{a} - \overset{\to}{b}\right) \times \left(\overset{\to}{a} + \overset{\to}{b}\right)\right|}^{2} + 4{\left(\overset{\to}{a} \cdot \overset{\to}{b}\right)}^{2} is equal to ______

Correct answer

0

Step-by-step solution

∣(a→−b→)×(a→+b→)∣2+4(a→⋅b→)2{\left|\left(\overset{\to}{a} - \overset{\to}{b}\right) \times \left(\overset{\to}{a} + \overset{\to}{b}\right)\right|}^{2} + 4{\left(\overset{\to}{a} \cdot \overset{\to}{b}\right)}^{2}

=∣a→×a→+a→×b→−b→×a→−b→×b→∣2+4(a→⋅b→)2= {\left|\overset{\to}{a} \times \overset{\to}{a} + \overset{\to}{a} \times \overset{\to}{b} - \overset{\to}{b} \times \overset{\to}{a} - \overset{\to}{b} \times \overset{\to}{b}\right|}^{2} + 4{\left(\overset{\to}{a} ⋅ \overset{\to}{b}\right)}^{2}

=4[∣a→×b→∣2+(a→⋅b→)2]=4∣a→∣2∣b→∣2=4×9×16=576= 4\left[{\left|\overset{\to}{a} \times \overset{\to}{b}\right|}^{2} + {\left(\overset{\to}{a} \cdot \overset{\to}{b}\right)}^{2}\right] = 4{\left|\overset{\to}{a}\right|}^{2}{\left|\overset{\to}{b}\right|}^{2} = 4 \times 9 \times 16 = 576

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