JEE Main 2022 · 25 Jul · Shift 1 · Mathematics Q15

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MathematicsSingle correct+4 / -1Question 15 of 90

Let ABC be a triangle such that $\overrightarrow{BC}=\vec a$ , $\overrightarrow{CA}=\vec b$ , $\overrightarrow{AB}=\vec c$ , $\left|\vec a\right|=6\sqrt2$ , $\left|\vec b\right|=2\sqrt3$ and $\vec b\cdot\vec c=12$ . Consider the statements :

(S1) : $\left|\left(\vec a\times\vec b\right)+\left(\vec c\times\vec b\right)\right|-\left|\vec c\right|=6\left(2\sqrt2-1\right)$

(S2) : $\angle ACB=\cos^{-1}\left(\sqrt{\dfrac23}\right)$

Then

both (S1) and (S2) are true

only (S1) is true

only (S2) is true

both (S1) and (S2) are false

Answer and worked solution

Official answer: C

Lengths. $\vec a+\vec b+\vec c=0$ gives $|\vec c|=6$ and $\vec a\cdot\vec b=-24$ .

S1. $(\vec a+\vec c)\times\vec b=-\vec b\times\vec b=0$ , so the value is $-6$ : false. S2. $\cos C=\frac{24}{12\sqrt6}=\sqrt{\frac23}$ : true.

(C)

Topics:Cross product Dot product

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JEE Main 2022 · 25 Jul · Shift 1, Mathematics question 15