If ${\bf{\vec a}},\,\,{\bf{\vec b}},\,\,{\bf{\vec c}}$ are non-coplanar vectors, then $\left| {\,\begin{array}{*{20}{c}} {{\bf{\vec a}} \cdot {\bf{\vec a}}}&{{\bf{\vec a}} \cdot {\bf{\vec b}}}&{{\bf{\vec a}} \cdot {\bf{\vec c}}}\\ {{\bf{\vec b}} \cdot {\bf{\vec a}}}&{{\bf{\vec b}} \cdot {\bf{\vec b}}}&{{\bf{\vec b}} \cdot {\bf{\vec c}}}\\ {{\bf{\vec c}} \cdot {\bf{\vec a}}}&{{\bf{\vec c}} \cdot {\bf{\vec b}}}&{{\bf{\vec c}} \cdot {\bf{\vec c}}} \end{array}\,} \right|$ is equal to

${\left[ {\;{\bf{\vec a}}\;\,{\bf{\vec b}}\;\,{\bf{\vec c}}\;} \right]^2}$

$\left[ {\;{\bf{\vec a}}\;\,{\bf{\vec b}}\;\,{\bf{\vec c}}\;} \right]$

${\left[ {\;{\bf{\vec a}}\,\;{\bf{\vec b}}\;\,{\bf{\vec c}}\;} \right]^{1/3}}$

JEE MATHEMATICS MCQ: SCALAR TRIPLE PRODUCT

1 Easy

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Understanding +4 | -1

\mathbf{\vec a},\,\,\mathbf{\vec b},\,\,\mathbf{\vec c}

are non-coplanar vectors, then

\left| {\,\begin{array}{cccccccccccccccccccc} {\mathbf{\vec a} \cdot \mathbf{\vec a}}&{\mathbf{\vec a} \cdot \mathbf{\vec b}}&{\mathbf{\vec a} \cdot \mathbf{\vec c}}\\ {\mathbf{\vec b} \cdot \mathbf{\vec a}}&{\mathbf{\vec b} \cdot \mathbf{\vec b}}&{\mathbf{\vec b} \cdot \mathbf{\vec c}}\\ {\mathbf{\vec c} \cdot \mathbf{\vec a}}&{\mathbf{\vec c} \cdot \mathbf{\vec b}}&{\mathbf{\vec c} \cdot \mathbf{\vec c}} \end{array}\,} \right|

is equal to

View answer and explanation

Correct option: A

{\left[ {\;\mathbf{\vec a}\;\,\mathbf{\vec b}\;\,\mathbf{\vec c}\;} \right]^2}

Option A (

{\left[ {\;\mathbf{\vec a}\;\,\mathbf{\vec b}\;\,\mathbf{\vec c}\;}} \right]^2}

) is correct because the determinant of the dot products is equivalent to the square of the scalar triple product.