JEE MATHEMATICS MCQ: CROSS PRODUCT

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Application +4 | -1

The moment of a force represented by

\mathbf{F}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}

about the point

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

is equal to

View answer and explanation

Correct option: C

-5 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}

Option C (

-5 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}

) is correct because it is the moment of the force about the given point, calculated using the cross product of the position vector and the force vector.

More Questions from CROSS PRODUCT

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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