JEE MATHEMATICS MCQ: CROSS PRODUCT

1 Easy

1 of 1

Application +4 | -1

The unit vector perpendicular to the vectors

\mathbf{i}-\mathbf{j}+\mathbf{k}

and

2 \mathbf{i}+3 \mathbf{j}-\mathbf{k}

View answer and explanation

Correct option: C

\frac{-2 i+3 \mathbf{i}+5 \mathbf{k}}{\sqrt{38}}

Option C (

\frac{{-2 i+3 \mathbf{j}+5 \mathbf{k}}}{{\sqrt{{38}}}}

) is correct because it is the unit vector obtained by dividing the cross product of the two vectors by its magnitude.

More Questions from CROSS PRODUCT

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

\vec{a}, \vec{b}

\lambda

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

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

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

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

S

O A

O C

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