Verified Solution • 2025–26 Session moderate

QJEEM20Maf8c5: JEE Main 2020 Physics - Class 11 JEEM20Maf8c5 Solution: Let , and be three unit vectors such that . If and , then the…

Let a→\overrightarrow a , b→\overrightarrow b and c→\overrightarrow c be three unit vectors such that
a→+b⃗+c→=0→\overrightarrow a + \vec b + \overrightarrow c = \overrightarrow 0 . If λ=a→.b⃗+b⃗.c→+c→.a→\lambda = \overrightarrow a .\vec b + \vec b.\overrightarrow c + \overrightarrow c .\overrightarrow a and
d→=a→×b⃗+b⃗×c→+c→×a→\overrightarrow d = \overrightarrow a \times \vec b + \vec b \times \overrightarrow c + \overrightarrow c \times \overrightarrow a , then the ordered pair, (λ,d→)\left( {\lambda ,\overrightarrow d } \right) is equal to :

Multiple Choice Options

  • A (32,3a→×c→)\left( {{3 \over 2},3\overrightarrow a \times \overrightarrow c } \right)
  • B (−32,3c→×b→)\left( { - {3 \over 2},3\overrightarrow c \times \overrightarrow b } \right)
  • C (−32,3a→×b→)\left( { - {3 \over 2},3\overrightarrow a \times \overrightarrow b } \right)
  • D (32,3b→×c→)\left( {{3 \over 2},3\overrightarrow b \times \overrightarrow c } \right)

Step-by-Step Verified Solution

a→+b⃗+c→=0→\overrightarrow a + \vec b + \overrightarrow c = \overrightarrow 0

⇒\Rightarrow ∣a→+b→+c→∣2{\left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right|^2} = 0

⇒\Rightarrow ∣a→∣2{{{\left| {\overrightarrow a } \right|}^2}} + ∣b→∣2{{{\left| {\overrightarrow b } \right|}^2}} + ∣c→∣2{{{\left| {\overrightarrow c } \right|}^2}} +
2(a→.b→){2\left( {\overrightarrow a .\overrightarrow b } \right)} + 2(b→.c→){2\left( {\overrightarrow b .\overrightarrow c } \right)} + 2(c→.a→){2\left( {\overrightarrow c .\overrightarrow {a} } \right)} = 0

⇒\Rightarrow 3 + 2(a→.b→+b→.c→+c→.a→){2\left( {\overrightarrow a .\overrightarrow b + \overrightarrow b .\overrightarrow c + \overrightarrow c .\overrightarrow a } \right)} = 0

⇒\Rightarrow 3 + 2 λ\lambda = 0

⇒\Rightarrow λ\lambda = −32- {3 \over 2}

d→=a→×b⃗+b⃗×c→+c→×a→\overrightarrow d = \overrightarrow a \times \vec b + \vec b \times \overrightarrow c + \overrightarrow c \times \overrightarrow a

= a→×b→+b→×(−a→−b→)+(−a→−b→)×a→{\overrightarrow a \times \overrightarrow b + \overrightarrow b \times \left( { - \overrightarrow a - \overrightarrow b } \right) + \left( { - \overrightarrow a - \overrightarrow b } \right) \times \overrightarrow a }

= a→×b→−b→×a→+{\overrightarrow a \times \overrightarrow b - \overrightarrow b \times \overrightarrow a + } b→×b→−a→×a→−b→×a→{\overrightarrow b \times \overrightarrow b - \overrightarrow a \times \overrightarrow a - \overrightarrow b \times \overrightarrow a }

= a→×b→+a→×b→+a→×b→{\overrightarrow a \times \overrightarrow b + \overrightarrow a \times \overrightarrow b + \overrightarrow a \times \overrightarrow b }

= 3(a→×b→){3\left( {\overrightarrow a \times \overrightarrow b } \right)}

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