If ${{\bf{\vec a}}}$ is vector perpendicular to both ${\bf{\vec b}}{\rm{\;and\;}}{\bf{\vec c}}$ then

${\bf{\vec a}} \times \left( {{\bf{\vec b}} \times {\bf{\vec c}}} \right) = \vec 0$

${\bf{\vec a}} + \left( {{\bf{\vec b}} + {\bf{\vec c}}} \right) = \vec 0$

${\bf{\vec a}} \times \left( {{\bf{\vec b}} + {\bf{\vec c}}} \right) = \vec 0$

${\bf{\vec a}}\cdot\left( {{\bf{\vec b}} \times {\bf{\vec c}}} \right) = \vec 0$

JEE MATHEMATICS MCQ: SCALAR TRIPLE PRODUCT

1 Easy

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Understanding +4 | -1

{\mathbf{\vec a}}

is vector perpendicular to both

\mathbf{\vec b}\mathrm{\;and\;}\mathbf{\vec c}

then

View answer and explanation

Correct option: C

\mathbf{\vec a} \times \left( {\mathbf{\vec b} \times \mathbf{\vec c}} \right) = \vec 0

Option B (

\mathbf{\vec a} \times ( {\mathbf{\vec b} + \mathbf{\vec c}} ) = \vec 0

) is correct because if

\mathbf{\vec a}

is perpendicular to both

\mathbf{\vec b}

and

\mathbf{\vec c}

, it is also perpendicular to their sum, and the cross product of perpendicular vectors is zero.

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}