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JEE Main 202313 AprilMorning Shift
Exam
JEE Main
Year
2023
Date
13 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Vector Algebra

JEE Main 2023 · 13 April · Morning Shift · Mathematics · Vector Algebra

Let → a = ^ i + 4 ^ j + 2 ^ k , → b = 3 ^ i - 2 ^ j + 7 ^ k and → c...

Let \overset{\to}{a} = \overset{^}{i} + 4\overset{^}{j} + 2\overset{^}{k}, \overset{\to}{b} = 3\overset{^}{i} - 2\overset{^}{j} + 7\overset{^}{k} and \overset{\to}{c} = 2\overset{^}{i} - \overset{^}{j} + 4\overset{^}{k}. If a vector d→\overset{\to}{d} satisfies d→×b→=c→×b→\overset{\to}{d} \times \overset{\to}{b} = \overset{\to}{c} \times \overset{\to}{b} and d→⋅a→=24,\overset{\to}{d} \cdot \overset{\to}{a} = 24, then ∣d→∣2{\left|\overset{\to}{d}\right|}^{2} is equal to

Options

  1. A323323
  2. B423423
  3. C313313
  4. D413413

Correct answer

D. 413413

Step-by-step solution

We have been given that

d→×b→=c→×b→\overset{\to}{d} \times \overset{\to}{b} = \overset{\to}{c} \times \overset{\to}{b}

⇒(d→−c→)×b→=0⇒ ( \overset{\to}{d} - \overset{\to}{c} ) \times \overset{\to}{b} = 0

⇒d→−c→=λb→(∵d→≠c→)⇒ \overset{\to}{d} - \overset{\to}{c} = \lambda\overset{\to}{b} ( ∵ \overset{\to}{d} \ne \overset{\to}{c} )

⇒d→=c→+λb→⇒ \overset{\to}{d} = \overset{\to}{c} + \lambda\overset{\to}{b}

It is given that a→⋅d→=24\overset{\to}{a} \cdot \overset{\to}{d} = 24

⇒a→⋅c→+λb→⋅a→=24⇒ \overset{\to}{a} \cdot \overset{\to}{c} + \lambda\overset{\to}{b} \cdot \overset{\to}{a} = 24

⇒6+λ(3−8+14)=24⇒ 6 + \lambda ( 3 - 8 + 14 ) = 24

⇒9λ=18⇒ 9\lambda = 18

⇒λ=2⇒ \lambda = 2

∴d→=c→+2b→∴ \overset{\to}{d} = \overset{\to}{c} + 2\overset{\to}{b}

⇒ \overset{\to}{d} = 8\overset{^}{i} - 5\overset{^}{j} + 18\overset{^}{k}

∣d→∣2=64+25+324| \overset{\to}{d}{|}^{2} = 64 + 25 + 324=413= 413

Hence, this is the correct option.

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