JEE Main 2020MathematicsVector AlgebraMediumNumerical

JEE Main 2020 — Vector Algebra Question with Solution

JEE Main 2020 (04 Sep Shift 2)

Question

If a→=2i^+j^+2k^, then, the value of i^×a→×i^2+j^×a→×j^2+k^×a→×k^2, is equal to :

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Show full solutionCorrect answer: 18
Correct answer
18

Step-by-step explanation

Let a→=xi^+yj^+zk^, and using vector triple product, we have,
i^×a→×i^=i^.i^a→-i^.a→i^=yj^+zk^
Similarly,

j^×a→×j^=xi^+zk^

and k^×a→×k^=xi^+yk^

Thus, 

i^×a→×i^2+j^×a→×j^2+k^×a→×k^2
=yj^+zk^2+xi^+zk^2+xi^+yj^2

=2|a→|2=2(9)=18

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About this question

This is a previous-year question from JEE Main 2020, covering the Vector Algebra chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.