JEE MATHEMATICS MCQ: VECTOR TRIPLE PRODUCT

1 Easy

1 of 1

Application +4 | -1

Let

\vec a = \hat i + \hat j - \hat k,\vec b = \hat i - \hat j + \hat k

and

{\vec c}

be a unit vector perpendicular to

{\vec a}

and coplanar with

{\vec a}

and

\vec b,

then it is given by

View answer and explanation

Correct option: A

\frac{1}{{\sqrt 6 }}\left( {2\hat i - \hat j + \hat k} \right)

Option A is correct because the unit vector perpendicular to

\vec{a}

and coplanar with

\vec{a}

and

\vec{b}

is found using the vector triple product and normalization.

More Questions from VECTOR TRIPLE PRODUCT

Practice All Chapter MCQs →

{\mathbf{\hat u}}

{\mathbf{\hat v}}

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

{\mathbf{\hat r}}

{\mathbf{\hat u}}

{\mathbf{\hat v}}

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

\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})