Let ${{\bf{\hat u}}}$ and ${{\bf{\hat v}}}$ are unit vectors such that ${\bf{\hat u}}\cdot{\bf{\hat v}} = 0$ If ${{\bf{\hat r}}}$ is any vector coplanar with ${{\bf{\hat u}}}$ and ${{\bf{\hat v}}}$, then the magnitude of the vector ${\bf{\vec r}} \times \left( {{\bf{\hat u}} \times {\bf{\hat v}}} \right)$ is

$2\left| {{\bf{\vec r}}} \right|$

JEE MATHEMATICS MCQ: VECTOR TRIPLE PRODUCT

1 Easy

1 of 1

Application +4 | -1

Let

{\mathbf{\hat u}}

and

{\mathbf{\hat v}}

are unit vectors such that

\mathbf{\hat u}\cdot\mathbf{\hat v} = 0

{\mathbf{\hat r}}

is any vector coplanar with

{\mathbf{\hat u}}

and

{\mathbf{\hat v}}

, then the magnitude of the vector

\mathbf{\vec r} \times \left( {\mathbf{\hat u} \times \mathbf{\hat v}} \right)

View answer and explanation

Correct option: C

\left| {\mathbf{\vec r}} \right|

Option C (

\left| {{\bf{{\vec r}}}} \right|

) is correct because the magnitude of the vector product simplifies to the magnitude of the original vector

{\bf{{\vec r}}}

Practice All Chapter MCQs →

\lambda \in \mathrm{R}, \vec{a}=\lambda \hat{i}+2 \hat{j}-3 \hat{k}, \vec{b}=\hat{i}-\lambda \hat{j}+2 \hat{k}

((\vec{a}+\vec{b}) \times(\vec{a} \times \vec{b})) \times(\vec{a}-\vec{b})=8 \hat{i}-40 \hat{j}-24 \hat{k}

|\lambda(\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})|^2

\vec{a}=-\hat{i}-\hat{j}+\hat{k}

\vec{a} \cdot \vec{b}=1

\vec{a} \times \vec{b}=\hat{i}-\hat{j}

\vec{a}-6 \vec{b}

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

\vec u = \vec a \times (\vec b \times \vec c),\vec v = \vec b \times (\vec c \times \vec a)\,\,\,{\text{and}}\,\,\,\vec w = \vec c \times (\vec a \times \vec b)

a

b

a\, \times (a\, \times (a\, \times (a\, \times b)))

\vec{a}

\hat{i} \times(\vec{a} \times \vec{i})+\hat{j} \times(\vec{a} \times \hat{j})+\hat{k} \times(\vec{a} \times \vec{k})

{a},{b}, {c}

{a} \cdot({b}+{c})={b} \cdot({c}+{a})={c} \cdot({a}+{b})=0

|{a}|=1,|{b}|=4

|{c}|=8

|{a}+{b}+{c}|=