The moment about the point $M\left( { - 2,\;4, - 6} \right)$ of the force represented in magnitude and position $\overrightarrow {{\bf{AB}}} $ where the points $A$ and $B$ have the coordinates (1, 2,$ - 3)$ and $\left( {3, - 4,2} \right)$ respectively, is

$8{\bf{\hat i}} - 9{\bf{\hat j}} - 14{\bf{\hat k}}$

$2{\bf{\hat i}} - 6{\bf{\hat j}} + 5{\bf{\hat k}}$

$ - 3{\bf{\hat i}} + 2{\bf{\hat j}} - 3{\bf{\hat k}}$

$ - 5{\bf{\hat i}} - 8{\bf{\hat j}} - 8{\bf{\hat k}}$

JEE MATHEMATICS MCQ: CROSS PRODUCT

1 Easy

1 of 1

Application +4 | -1

The moment about the point

M\left( { - 2,\;4, - 6} \right)

of the force represented in magnitude and position

\overrightarrow {\mathbf{AB}}

where the points

A

and

B

have the coordinates (1, 2,

- 3)

and

\left( {3, - 4,2} \right)

respectively, is

View answer and explanation

Correct option: A

8\mathbf{\hat i} - 9\mathbf{\hat j} - 14\mathbf{\hat k}

The correct answer is A. The moment is calculated by taking the cross product of the position vector from M to A and the force vector AB.

Practice All Chapter MCQs →

\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}