JEE MATHEMATICS MCQ: CROSS PRODUCT

1 Medium

1 of 1

Application +4 | -1

A unit vector perpendicular to the plane determinal by the points

{P}(1,-1,2), {Q}(2,0,-1)

and

{R}(0,2,1)

View answer and explanation

Correct option: A

\frac{2 {i}+{j}+{k}}{\sqrt{6}}

Option A (

\frac{2 {i}+{j}+{k}}{\sqrt{6}}

) is correct because it represents the unit vector perpendicular to the plane determined by the given points, found by taking the cross product of two vectors in the plane and normalizing.

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}

, where

\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

for some real

x.

Then

\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

and

c

30^\circ ,

then

|(a \times b) \times c|

equal to If

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

, then

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

is Let

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

, where

O

is the origin, If

S

is the parallelogram with adjacent sides

O A

and

O C

, then

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

is equal to