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JEE MATHEMATICS MCQ: CROSS PRODUCT

1 Easy

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Application +4 | -1

If

\vec{a} \perp \vec{b}

, then vector

\vec{v}

in terms of

\vec{a}

and

\vec{b}

satisfying the condition

\vec{v} \cdot \vec{a}

and

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

and

\left[ {\begin{array}{cccccccccccccccccccc} {\vec v}&{\vec a}&{\vec b} \end{array}} \right] = 1

is

View answer and explanation

Correct option: A

\frac{\overrightarrow{\mathrm{b}}}{|\overrightarrow{\mathrm{b}}|^{2}}+\frac{\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}}{|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|^{2}}

Option A (

\frac{{\overrightarrow{{\mathrm{{b}}}}}}{{\|\overrightarrow{{\mathrm{{b}}}}\|^2}} + \frac{{\overrightarrow{{\mathrm{{a}}}} \times \overrightarrow{{\mathrm{{b}}}}}}{{\|\overrightarrow{{\mathrm{{a}}}} \times \overrightarrow{{\mathrm{{b}}}}\|^2}}

) is correct because it satisfies the given conditions

\vec{v} \cdot \vec{a} = 0

,

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

, and

\left[ {\begin{{array}}{{*{20}{c}}} {\vec v}&{\vec a}&{\vec b} \end{{array}}} \right] = 1

.

More Questions from CROSS PRODUCT

Practice All Chapter MCQs →

A solution of the vector equation

\vec{r} \times \vec{b}=\vec{a} \times \vec{b}

\vec{a}, \vec{b}

are two given vectors, is where

\lambda

is a parameter. Let

a = 3\hat i + 2\hat j + x\hat k\mathrm{ }\,\mathrm{\,\,and}\,\,\mathrm{ }b = \hat i - \hat j + \hat k

x.

\left| {a \times b} \right| = r

is possible if A tetrahedron has vertices P(1, 2, 1), Q(2, 1, 3), R(-1, 1, 2) and O(0, 0, 0). The angle between the faces OPQ and PQR is Let

a = 2\widehat i + \widehat j - 2\widehat k{\text{ }}\,\,{\text{and }}\,\,b = \widehat i + \widehat j.

If c is a vector such that

a \cdot c = |c|,\,\,|c - a\mid = 2\sqrt 2

and the angle between

a \times b

c

30^\circ ,

|(a \times b) \times c|

\vec{A}=i-2 j-3 k, \vec{B}=2 i+j-k, \vec{C}=i+3 j-2 k

(\vec{A} \times \vec{B}) \times \vec{C}

\overrightarrow{O A}=\vec{a}, \overrightarrow{O B}=12 \vec{a}+4 \vec{b}

O

is the origin, If

S

is the parallelogram with adjacent sides

O A

O C

\frac{\text { area of the quadrilateral } O A B C}{\text { area of } S}