$\left[ {{\bf{\vec b}} \times {\bf{\vec c}}\;\;{\bf{\vec c}} \times {\bf{\vec a}}\;\;{\bf{\vec a}} \times {\bf{\vec b}}} \right]$ is equal to

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

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

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

${\bf{\vec a}}\; \times \left( {{\bf{\vec b}} \times {\bf{\vec c}}} \right)$

JEE MATHEMATICS MCQ: SCALAR TRIPLE PRODUCT

1 Easy

1 of 1

Analyzing +4 | -1

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

is equal to

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because the scalar triple product of cross products simplifies to the square of the original scalar triple product.

Practice All Chapter MCQs →

\vec r\cdot\vec a = \vec r\cdot\vec b = \vec r\cdot\vec c = 0

\vec r,

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

\frac{{x - 3}}{2} = \frac{{y - 2}}{3} = \frac{{z - 1}}{{ - 1}}

\frac{{x + 3}}{2} = \frac{{y - 6}}{1} = \frac{{z - 5}}{3},

\mathbf{\vec a} = \widehat \mathbf{i} - \mathbf{\hat k},\;\mathbf{\vec b} = x\widehat \mathbf{i} + \widehat \mathbf{j} + \left( {1 - x} \right)\mathbf{\hat k}

\mathbf{\vec c} = y\widehat \mathbf{i} + x\widehat \mathbf{j} + \left( {1 + x - y} \right)\mathbf{\hat k}.

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

\vec{a}, \vec{b}

((\vec{a} \times \vec{b}) \times \vec{a}) \cdot((\vec{b} \times \vec{a}) \times \vec{b})

a \cdot(a \times b)=

\mathbf{r}=\mathbf{b}+\lambda \mathbf{c}