JEE MATHEMATICS MCQ: VECTOR TRIPLE PRODUCT

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The vector (s) which is (are) coplanar with vectors

\hat{i}+\hat{j}+2 \hat{k}

and

\hat{i}+2 \hat{j}+\hat{k},

and perpendicular to the vector

\hat{i}+\hat{j}+\hat{k},

is/are

View answer and explanation

Correct option: A

\hat{j}-\hat{k}

and

-\hat{j}+\hat{k}

Option B (

\frac{\sqrt{17}}{9}

) is correct because it is the cosine of the angle between the original vector and the rotated vector, calculated using the dot product and magnitudes.

More Questions from VECTOR TRIPLE PRODUCT

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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)

is Let

\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}

then

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

is equal to Let

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

such that

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

and

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

then

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

equals Let

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

be any three vectors. Then, vectors

\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)

are such that they are If

a

and

b

are perpendicular, then

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

is equal to For any vector

\vec{a}

, the value of

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

, is