Verified Solution • 2025–26 Session moderate

QJEEM20Maecdb: JEE Main 2020 Physics - Class 11 JEEM20Maecdb Solution: Let and be two vectors. If is a vector such that and , then is…

Let a→=i^−2j^+k^\overrightarrow a = \widehat i - 2\widehat j + \widehat k and b→=i^−j^+k^\overrightarrow b = \widehat i - \widehat j + \widehat k be two vectors. If c→\overrightarrow c is a vector such that b→×c→=b→×a→\overrightarrow b \times \overrightarrow c = \overrightarrow b \times \overrightarrow a and c→.a→=0\overrightarrow c .\overrightarrow a = 0 , then c→.b→\overrightarrow c .\overrightarrow b is equal to

Multiple Choice Options

  • A −12- {1 \over 2}
  • B −32- {3 \over 2}
  • C 12{1 \over 2}
  • D -1

Step-by-Step Verified Solution

a→=i^−2j^+k^\overrightarrow a = \widehat i - 2\widehat j + \widehat k

b→=i^−j^+k^\overrightarrow b = \widehat i - \widehat j + \widehat k

∣a→∣\left| {\overrightarrow a } \right| = 6\sqrt 6 , ∣b→∣\left| {\overrightarrow b } \right| = 3\sqrt 3

and a→.b→{\overrightarrow a .\overrightarrow b } = 4

Given b→×c→=b→×a→\overrightarrow b \times \overrightarrow c = \overrightarrow b \times \overrightarrow a

⇒\Rightarrow (b→×c→){\left( {\overrightarrow b \times \overrightarrow c } \right)} - (b→×a→){\left( {\overrightarrow b \times \overrightarrow a } \right)} = 0

⇒\Rightarrow b→×(c→−a→){\overrightarrow b \times \left( {\overrightarrow c - \overrightarrow a } \right)} = 0

∴\therefore b→∥(c→−a→){\overrightarrow b \parallel \left( {\overrightarrow c - \overrightarrow a } \right)}

⇒\Rightarrow (c→−a→)=λb→{\left( {\overrightarrow c - \overrightarrow a } \right) = \lambda \overrightarrow b }

⇒\Rightarrow c→=a→+λb→{\overrightarrow c = \overrightarrow a + \lambda \overrightarrow b }

⇒\Rightarrow c→.a→=a→.a→+λa→.b→{\overrightarrow c .\overrightarrow a = \overrightarrow a .\overrightarrow a + \lambda \overrightarrow a .\overrightarrow b }

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

⇒\Rightarrow λ\lambda = −∣a→∣2a→.b→{{{ - {{\left| {\overrightarrow a } \right|}^2}} \over {\overrightarrow a .\overrightarrow b }}} = −64=−32{{ - 6} \over 4} = - {3 \over 2}

∴\therefore c→\overrightarrow c = a→−32b→{\overrightarrow a - {3 \over 2}\overrightarrow b }

⇒\Rightarrow c→\overrightarrow c = ( i^−2j^+k^\widehat i - 2\widehat j + \widehat k ) - 32{3 \over 2} ( i^−j^+k^\widehat i - \widehat j + \widehat k )

⇒\Rightarrow c→\overrightarrow c = −12(i^+j^+k^)- {1 \over 2}\left( {\widehat i + \widehat j + \widehat k} \right)

∴\therefore c→.b→\overrightarrow c .\overrightarrow b = −12(i^+j^+k^)- {1 \over 2}\left( {\widehat i + \widehat j + \widehat k} \right) ( i^−j^+k^\widehat i - \widehat j + \widehat k )

∴\therefore c→.b→\overrightarrow c .\overrightarrow b = −12- {1 \over 2}

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