JEE Main 2020 · Mathematics · Vector Algebra · medium

Let vec a,vec b be unit vectors inclined at an angle of 60^°. The volume of the parallelepiped whose coterminous edges are vec a, vec b and vec a×vec…

Single correct option

Let

\vec a,\vec b

be unit vectors inclined at an angle of

60^\circ

. The volume of the parallelepiped whose coterminous edges are

\vec a

\vec b

and

\vec a\times\vec b

Show answer and solution

Answer: (C) $\dfrac34$

The volume is the modulus of the scalar triple product:

V=\big|[\vec a\ \ \vec b\ \ \vec a\times\vec b]\big|=\big|(\vec a\times\vec b)\cdot(\vec a\times\vec b)\big|=|\vec a\times\vec b|^2,

because

[\vec a\,\vec b\,\vec c]=\vec c\cdot(\vec a\times\vec b)

with

\vec c=\vec a\times\vec b

. Now

|\vec a\times\vec b|=|\vec a||\vec b|\sin60^\circ=\dfrac{\sqrt3}{2}

, hence

V=\dfrac34

. The value

\dfrac{\sqrt3}{2}

is only the area of the base parallelogram.

Original question written in the JEE Main 2020 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake