Verified Solution • 2025–26 Session moderate

QJEEM21Ma92cf: JEE Main 2021 Physics - Class 11 JEEM21Ma92cf Solution: Let and be two non-zero vectors perpendicular to each other and …

Let a→\overrightarrow a and b→\overrightarrow b be two non-zero vectors perpendicular to each other and ∣a→∣=∣b→∣|\overrightarrow a | = |\overrightarrow b | . If ∣a→×b→∣=∣a→∣|\overrightarrow a \times \overrightarrow b | = |\overrightarrow a | , then the angle between the vectors (a→+b→+(a→×b→))\left( {\overrightarrow a + \overrightarrow b + \left( {\overrightarrow a \times \overrightarrow b } \right)} \right) and a→{\overrightarrow a } is equal to :

Multiple Choice Options

  • A sin⁡−1(16){\sin ^{ - 1}}\left( {{1 \over {\sqrt 6 }}} \right)
  • B cos⁡−1(12){\cos ^{ - 1}}\left( {{1 \over {\sqrt 2 }}} \right)
  • C sin⁡−1(13){\sin ^{ - 1}}\left( {{1 \over {\sqrt 3 }}} \right)
  • D cos⁡−1(13){\cos ^{ - 1}}\left( {{1 \over {\sqrt 3 }}} \right)

Step-by-Step Verified Solution

a→\overrightarrow a is perpendicular to b→\overrightarrow b

∴\therefore a→\overrightarrow a . b→\overrightarrow b = 0

Given, | a→\overrightarrow a ×\times b→\overrightarrow b | = | a→\overrightarrow a |

and | a→\overrightarrow a | = | b→\overrightarrow b |

∴\therefore | a→\overrightarrow a ×\times b→\overrightarrow b | = | a→\overrightarrow a | = | b→\overrightarrow b | = k(assume)

Now, angle between a→\overrightarrow a and ( a→\overrightarrow a + b→\overrightarrow b + ( a→\overrightarrow a ×\times b→\overrightarrow b ))

cos⁡θ=a→((a→+b→+(a→×b→))∣a→∣.∣a→+b→+(a→×b→)∣\cos \theta = {{\overrightarrow a ((\overrightarrow a + \overrightarrow b + (\overrightarrow a \times \overrightarrow b ))} \over {|\overrightarrow a |.|\overrightarrow a + \overrightarrow b + (\overrightarrow a \times \overrightarrow b )|}}

=∣a→∣2+ a→. b→+a→(a→×b→)∣a→∣.∣a→+b→+(a→×b→)∣= {{|\overrightarrow a {|^2} + \,\overrightarrow a .\,\overrightarrow b + \overrightarrow a (\overrightarrow a \times \overrightarrow b )} \over {|\overrightarrow a |.|\overrightarrow a + \overrightarrow b + (\overrightarrow a \times \overrightarrow b )|}}

[Note a→.(a→×b→)=[a→a→b→]=0\overrightarrow a .(\overrightarrow a \times \overrightarrow b ) = [\overrightarrow a \overrightarrow a \overrightarrow b ] = 0 ]

=∣a→∣2+0+0∣a→∣.∣a→+b→+(a→×b→)∣= {{|\overrightarrow a {|^2} + 0 + 0} \over {|\overrightarrow a |.|\overrightarrow a + \overrightarrow b + (\overrightarrow a \times \overrightarrow b )|}}

Now, ∣a→+b→+(a→×b→)∣2|\overrightarrow a + \overrightarrow b + (\overrightarrow a \times \overrightarrow b ){|^2}

=∣a→∣2+∣a→∣2+∣a→×b→∣2+= |\overrightarrow a {|^2} + |\overrightarrow a {|^2} + |\overrightarrow a \times \overrightarrow b {|^2} +

2a→.b→+2b→.(a→×b→)+2a→.(a→×b→)2\overrightarrow a .\overrightarrow b + 2\overrightarrow b .(\overrightarrow a \times \overrightarrow b ) + 2\overrightarrow a .(\overrightarrow a \times \overrightarrow b )

=k2+k2+k2= {k^2} + {k^2} + {k^2}

∴\therefore ∣a→+b→+(a→×b→)∣2=3k2|\overrightarrow a + \overrightarrow b + (\overrightarrow a \times \overrightarrow b ){|^2} = 3{k^2}

⇒∣a→+b→+(a→×b→)∣=3k\Rightarrow |\overrightarrow a + \overrightarrow b + (\overrightarrow a \times \overrightarrow b )| = \sqrt 3 k

∴\therefore cos⁡θ=k2k(3k)=13\cos \theta = {{{k^2}} \over {k(\sqrt 3 k)}} = {1 \over {\sqrt 3 }}

⇒θ=cos⁡−1(13)\Rightarrow \theta = {\cos ^{ - 1}}\left( {{1 \over {\sqrt 3 }}} \right)

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