JEE Main 2022 · 29 Jul · Shift 2 · Mathematics Q18

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MathematicsSingle correct+4 / -1Question 18 of 90

Let $\vec a,\vec b,\vec c$ be three coplanar concurrent vectors such that angles between any two of them is same. If the product of their magnitudes is 14 and $(\vec a\times\vec b)\cdot(\vec b\times\vec c)+(\vec b\times\vec c)\cdot(\vec c\times\vec a)+(\vec c\times\vec a)\cdot(\vec a\times\vec b)=168$ , then $|\vec a|+|\vec b|+|\vec c|$ is equal to :

Answer and worked solution

Official answer: C

Angles are all 120°. Each term, e.g. $(\vec a\times\vec b)\cdot(\vec b\times\vec c)=(\vec a\cdot\vec b)(\vec b\cdot\vec c)-(\vec a\cdot\vec c)|\vec b|^2=\frac34|\vec a||\vec b|^2|\vec c|$ .

(C)

Topics:Cross product Dot product

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JEE Main 2022 · 29 Jul · Shift 2, Mathematics question 18