If ${\bf{\vec a}},\;{\bf{\vec b}},\;{\bf{\vec c}}$ are three non-coplanar vectors, then $\left( {{\bf{\vec a}} + {\bf{\vec b}} + {\bf{\vec c}}} \right)\cdot\left[ {\left( {{\bf{\vec a}} + {\bf{\vec b}}} \right) \times \left( {{\bf{\vec a}} \times {\bf{\vec c}}} \right)} \right]$ is

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

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

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

JEE MATHEMATICS MCQ: SCALAR TRIPLE PRODUCT

1 Easy

1 of 1

Analyzing +4 | -1

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

are three non-coplanar vectors, then

\left( {\mathbf{\vec a} + \mathbf{\vec b} + \mathbf{\vec c}} \right)\cdot\left[ {\left( {\mathbf{\vec a} + \mathbf{\vec b}} \right) \times \left( {\mathbf{\vec a} \times \mathbf{\vec c}} \right)} \right]

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because after expanding the scalar triple product, the result simplifies to the negative of the scalar triple product of a, b, and c.