Verified Solution • 2025–26 Session moderate

QJEEM22Ma38e1: JEE Main 2022 Physics - Class 11 JEEM22Ma38e1 Solution: Let , be unit vectors. If be a vector such that the angle…

Let a^\widehat a , b^\widehat b be unit vectors. If c→\overrightarrow c be a vector such that the angle between a^\widehat a and c→\overrightarrow c is π12{\pi \over {12}} , and b^=c→+2(c→×a^)\widehat b = \overrightarrow c + 2\left( {\overrightarrow c \times \widehat a} \right) , then ∣6c→∣2{\left| {6\overrightarrow c } \right|^2} is equal to :

Multiple Choice Options

  • A 6(3−3)6\left( {3 - \sqrt 3 } \right)
  • B 3+33 + \sqrt 3
  • C 6(3+3)6\left( {3 + \sqrt 3 } \right)
  • D 6(3+1)6\left( {\sqrt 3 + 1} \right)

Step-by-Step Verified Solution

∵b^=c⃗+2(c⃗×a^)\because \quad \hat{b}=\vec{c}+2(\vec{c} \times \hat{a})

amp;⇒b^⋅c⃗=∣c⃗∣2amp;∴b^−c⃗=2(c⃗×a⃗)\begin{aligned}&\Rightarrow \hat{b} \cdot \vec{c}=|\vec{c}|^{2} \\\\&\therefore \hat{b}-\vec{c}=2(\vec{c} \times \vec{a})\end{aligned}

amp;⇒∣b^∣2+∣c⃗∣2−2b^⋅c⃗=4∣c⃗∣2∣a⃗∣2sin⁡2π12amp;⇒1+∣c⃗∣2−2∣c∣2=4∣c⃗∣2(3−122)2amp;⇒1=∣c⃗∣2(3−3)amp;⇒36∣c⃗∣2=363−3=6(3+3)\begin{aligned}&\Rightarrow|\hat{b}|^{2}+|\vec{c}|^{2}-2 \hat{b} \cdot \vec{c}=4|\vec{c}|^{2}|\vec{a}|^{2} \sin ^{2} \frac{\pi}{12} \\\\&\Rightarrow 1+|\vec{c}|^{2}-2|c|^{2}=4|\vec{c}|^{2}\left(\frac{\sqrt{3}-1}{2 \sqrt{2}}\right)^{2} \\\\&\Rightarrow 1=|\vec{c}|^{2}(3-\sqrt{3}) \\\\&\Rightarrow 36|\vec{c}|^{2}=\frac{36}{3-\sqrt{3}}=6(3+\sqrt{3})\end{aligned}

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring