For any three vectors ${\bf{\vec a}},{\bf{\vec b}}$ and ${{\bf{\vec c}}}$, $\left( {{\bf{\vec a}} - {\bf{\vec b}}} \right)\cdot\left( {{\bf{\vec b}} + {\bf{\vec c}}\;} \right) \times \left( {\;{\bf{\vec c}} + {\bf{\vec a}}} \right)$ is equal to

$2{\bf{\vec a}}\cdot\left( {{\bf{\vec b}} \times {\bf{\vec c}}\;} \right)$

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

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

JEE MATHEMATICS MCQ: SCALAR TRIPLE PRODUCT

1 Easy

1 of 1

Analyzing +4 | -1

For any three vectors

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

and

{\mathbf{\vec c}}

\left( {\mathbf{\vec a} - \mathbf{\vec b}} \right)\cdot\left( {\mathbf{\vec b} + \mathbf{\vec c}\;} \right) \times \left( {\;\mathbf{\vec c} + \mathbf{\vec a}} \right)

is equal to

View answer and explanation

Correct option: D

Option D (0) is correct because the scalar triple product simplifies to zero due to vector properties and cancellations.

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}