If the vectors ${\bf{\vec c}},{\bf{\vec a}} = x\widehat {\bf{i}} + y\widehat {\bf{j}} + z{\bf{\hat k}}$ and ${\bf{\vec b}} = \widehat {\bf{j}}$ are such that ${\bf{\vec a}},{\bf{\vec c}}{\rm{\;and\;}}{\bf{\vec b}}$ from a right handed system, then ${{\bf{\vec c}}}$ is

$z\widehat {\bf{i}} - x{\bf{\hat k}}$

$ - z\widehat {\bf{i}} - x{\bf{\hat k}}$

JEE MATHEMATICS MCQ: SCALAR TRIPLE PRODUCT

1 Easy

1 of 1

Application +4 | -1

If the vectors

\mathbf{\vec c},\mathbf{\vec a} = x\widehat \mathbf{i} + y\widehat \mathbf{j} + z\mathbf{\hat k}

and

\mathbf{\vec b} = \widehat \mathbf{j}

are such that

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

from a right handed system, then

{\mathbf{\vec c}}

View answer and explanation

Correct option: A

z\widehat \mathbf{i} - x\mathbf{\hat k}

Option A (

z\widehat \mathbf{i} - x\mathbf{\hat k}

) is correct because in a right-handed system, the cross product of \vec{b} and \vec{a} gives \vec{c}.

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}