JEE Main 2022 · MathsMultiple choiceSingle correctHardMulti-step

JEE Main 29 July 2022, Shift 2, Maths Q18

Question 18 of 90 in this shift, Maths question 18 of 30, Section A.

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 :

(1) $10$
(2) $14$
(3) $16$ Official answer
(4) $18$

Official answer

Option 3

NTA final key.

Chapter: Vector Algebra
Topic: Cross Product
Idea tested: Lagrange's Identity

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Asked 8× in all

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Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.