JEE MATHEMATICS MCQ: VECTOR TRIPLE PRODUCT

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Analyzing +4 | -1

Let

a = \widehat i - \widehat j,\,\,\,b = \widehat j - \widehat k

and

c = \widehat k - \widehat i.

If d is a unit vectors such that

a \cdot d = 0 = (bcd),

then d is equal to

View answer and explanation

Correct option: A

\pm \frac{{(\widehat i + \widehat j - 2\widehat k)}}{{\sqrt 6 }}

The correct answer is A (

\pm \frac{{(\widehat i + \widehat j - 2\widehat k)}}{{{\sqrt 6 }}}

) because it is derived by finding a vector perpendicular to a and coplanar with b and c, then normalizing it to obtain the unit vector d.

More Questions from VECTOR TRIPLE PRODUCT

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{\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})