If ${\bf{\vec a}}{\rm{\;}}$ is perpendicular to ${{\bf{\vec b}}}$ and ${\bf{\vec c}}\left| {{\bf{\vec a}}} \right| = 2,\;\left| {{\bf{\vec b}}\left| { = 3,\;} \right|{\bf{\vec c}}} \right| = 4$ and the angle between ${{\bf{\vec b}}}$ and ${\bf{\vec c}}\;{\rm{is}}\frac{{2{\rm{\pi }}}}{3}$, then $\left[ {{\bf{\vec a}}\;{\bf{\vec b}}\;{\bf{\vec c}}} \right]$ is equal to

JEE MATHEMATICS MCQ: SCALAR TRIPLE PRODUCT

1 Easy

1 of 1

Application +4 | -1

\mathbf{\vec a}\mathrm{\;}

is perpendicular to

{\mathbf{\vec b}}

and

\mathbf{\vec c}\left| {\mathbf{\vec a}} \right| = 2,\;\left| {\mathbf{\vec b}\left| { = 3,\;} \right|\mathbf{\vec c}} \right| = 4

and the angle between

{\mathbf{\vec b}}

and

\mathbf{\vec c}\;\mathrm{is}\frac{{2\mathrm{\pi }}}{3}

, then

\left[ {\mathbf{\vec a}\;\mathbf{\vec b}\;\mathbf{\vec c}} \right]

is equal to

View answer and explanation

Correct option: C

12\sqrt 3

Option C (

12\sqrt 3

) is correct because the scalar triple product is calculated using the magnitudes of the vectors and the sine of the angle between them.