Verified Solution • 2025–26 Session moderate

QJEEM18Ma16bf: JEE Main 2018 Physics - Class 11 JEEM18Ma16bf Solution: Let and a vector be such that and Then equals :

Let a→=i^+j^+k^,c→=j^−k^\overrightarrow a = \widehat i + \widehat j + \widehat k,\overrightarrow c = \widehat j - \widehat k and a vector b→\overrightarrow b be such that a→×b→=c→\overrightarrow a \times \overrightarrow b = \overrightarrow c and a→.b→=3.\overrightarrow a .\overrightarrow b = 3. Then ∣b→∣\left| {\overrightarrow b } \right| equals :

Multiple Choice Options

  • A 113{{11} \over 3}
  • B 113{{11} \over {\sqrt 3 }}
  • C 113\sqrt {{{11} \over 3}}
  • D 113{{\sqrt {11} } \over 3}

Step-by-Step Verified Solution

∵\because   a→\overrightarrow a == i^+j^+k^⇒∣a→∣=3\widehat i + \widehat j + \widehat k \Rightarrow \left| {\overrightarrow a } \right| = \sqrt 3

&   c→=j^−k^⇒∣c→∣2\overrightarrow c = \widehat j - \widehat k \Rightarrow \left| {\overrightarrow c } \right|\sqrt 2

Now, a→\overrightarrow a ×\times b→\overrightarrow b = c→\overrightarrow c     (Given)

⇒\Rightarrow   ∣a⃗∣∣b⃗∣sin⁡θ=∣c→∣\left| {\vec a} \right|\left| {\vec b} \right|\sin \theta = \left| {\overrightarrow c } \right|

⇒\Rightarrow   ∣a→∣∣b→∣sin⁡θ=2\left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\sin \theta = \sqrt 2

also   a→.b→=3\overrightarrow a .\overrightarrow b = 3

⇒\Rightarrow   ∣a→∣∣b→∣cos⁡θ=3\left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\cos \theta = 3

Dividing [i] by [iii], we get

tan θ\theta = 23{{\sqrt 2 } \over 3}

∴\therefore sin θ\theta == 211{{\sqrt 2 } \over {\sqrt {11} }}

Substituting value of sin θ\theta in [i] we get

3∣b→∣211=2\sqrt 3 \left| {\overrightarrow b } \right|{{\sqrt 2 } \over {\sqrt {11} }} = \sqrt 2

∣b→∣=113\left| {\overrightarrow b } \right| = {{\sqrt {11} } \over {\sqrt 3 }}

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