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JEE MAIN - Mathematics (2022 - 25th July Morning Shift - No. 14)

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

$$(\mathrm{S} 1):|(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})+(\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{b}})|-|\vec{c}|=6(2 \sqrt{2}-1)$$

$$(\mathrm{S} 2): \angle \mathrm{ACB}=\cos ^{-1}\left(\sqrt{\frac{2}{3}}\right)$$

Then

both (S1) and (S2) are true
only (S1) is true
only (S2) is true
both (S1) and (S2) are false

Explanation

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

$$ \because \vec{a}+\vec{b}+\vec{c}=0 $$

then $\bar{a}+\vec{c}=-\vec{b}$

then $(\vec{a}+\vec{c}) \times \vec{b}=-\vec{b} \times \bar{b}$

$\therefore \quad \vec{a} \times \vec{b}+\vec{c} \times \vec{b}=\overline{0}\quad\dots(i)$

For $(S 1):|\vec{a} \times \vec{b}+\vec{c} \times \vec{b}|-|\vec{c}|=6(2 \sqrt{2}-1)$

$$ \begin{aligned} &|(\vec{a}+\vec{c}) \times \vec{b}|-|\vec{c}|=6(2 \sqrt{2}-1) \\\\ &|\vec{c}|=6-12 \sqrt{2} \text { (not possible) } \end{aligned} $$

Hence (S1) is not correct

For (S2) : from (i) $\vec{b}+\vec{c}=-\vec{a}$

$\Rightarrow \vec{b} \cdot \vec{b}+\vec{c} \cdot \vec{b}=-\vec{a} \cdot \vec{b}$

$\Rightarrow 12+12=-6 \sqrt{2} \cdot 2 \sqrt{3} \cos (\pi-\angle A C B)$

$\therefore \cos (\angle A C B)=\sqrt{\frac{2}{3}}$

$$ \begin{aligned} &\therefore \angle A C B=\cos ^{-1} \sqrt{\frac{2}{3}} \\\\ &\therefore S(2) \text { is correct. } \end{aligned} $$

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