Verified Solution • 2025–26 Session moderate

QJEEM22Maefff: JEE Main 2022 Physics - Class 11 JEEM22Maefff Solution: Let be a triangle such that and . Consider the statements : Then

Let ABC\mathrm{ABC} be a triangle such that BC→=a→,CA→=b→,AB→=c→,∣a→∣=62,∣b→∣=23\overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{CA}}=\overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{c}},|\overrightarrow{\mathrm{a}}|=6 \sqrt{2},|\overrightarrow{\mathrm{b}}|=2 \sqrt{3} and b⃗⋅c⃗=12\vec{b} \cdot \vec{c}=12 . Consider the statements :

(S1):∣(a→×b→)+(c→×b→)∣−∣c⃗∣=6(22−1)(\mathrm{S} 1):|(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})+(\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{b}})|-|\vec{c}|=6(2 \sqrt{2}-1)

(S2):∠ACB=cos⁡−1(23)(\mathrm{S} 2): \angle \mathrm{ACB}=\cos ^{-1}\left(\sqrt{\frac{2}{3}}\right)

Then

Multiple Choice Options

  • A both (S1) and (S2) are true
  • B only (S1) is true
  • C only (S2) is true
  • D both (S1) and (S2) are false

Step-by-Step Verified Solution

JEE Main 2022 (Online) 25th July Morning Shift Mathematics - Vector Algebra Question 111 English Explanation

∵a⃗+b⃗+c⃗=0\because \vec{a}+\vec{b}+\vec{c}=0

then aˉ+c⃗=−b⃗\bar{a}+\vec{c}=-\vec{b}

then (a⃗+c⃗)×b⃗=−b⃗×bˉ(\vec{a}+\vec{c}) \times \vec{b}=-\vec{b} \times \bar{b}

∴a⃗×b⃗+c⃗×b⃗=0‾…(i)\therefore \quad \vec{a} \times \vec{b}+\vec{c} \times \vec{b}=\overline{0}\quad\dots(i)

For (S1):∣a⃗×b⃗+c⃗×b⃗∣−∣c⃗∣=6(22−1)(S 1):|\vec{a} \times \vec{b}+\vec{c} \times \vec{b}|-|\vec{c}|=6(2 \sqrt{2}-1)

amp;∣(a⃗+c⃗)×b⃗∣−∣c⃗∣=6(22−1)amp;∣c⃗∣=6−122 (not possible) \begin{aligned}&|(\vec{a}+\vec{c}) \times \vec{b}|-|\vec{c}|=6(2 \sqrt{2}-1) \\\\&|\vec{c}|=6-12 \sqrt{2} \text { (not possible) }\end{aligned}

Hence (S1) is not correct

For (S2) : from (i) b⃗+c⃗=−a⃗\vec{b}+\vec{c}=-\vec{a}

⇒b⃗⋅b⃗+c⃗⋅b⃗=−a⃗⋅b⃗\Rightarrow \vec{b} \cdot \vec{b}+\vec{c} \cdot \vec{b}=-\vec{a} \cdot \vec{b}

⇒12+12=−62⋅23cos⁡(π−∠ACB)\Rightarrow 12+12=-6 \sqrt{2} \cdot 2 \sqrt{3} \cos (\pi-\angle A C B)

∴cos⁡(∠ACB)=23\therefore \cos (\angle A C B)=\sqrt{\frac{2}{3}}

amp;∴∠ACB=cos⁡−123amp;∴S(2) is correct. \begin{aligned}&\therefore \angle A C B=\cos ^{-1} \sqrt{\frac{2}{3}} \\\\&\therefore S(2) \text { is correct. }\end{aligned}

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