The moment about the point M(-2, 4, -6) of th | vedclass

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(A) The force vector $\overrightarrow{F}$ is given by $\overrightarrow{AB} = (3 - 1)i + (-4 - 2)j + (2 - (-3))k = 2i - 6j + 5k$.The moment of force $\

Vedclass · Accepted Answer

$8i - 9j - 14k$

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(A) The force vector $\overrightarrow{F}$ is given by $\overrightarrow{AB} = (3 - 1)i + (-4 - 2)j + (2 - (-3))k = 2i - 6j + 5k$.The moment of force $\overrightarrow{F}$ about point $M$ is given by $\overrightarrow{	au} = \overrightarrow{MA} 	imes \overrightarrow{F}$.First,calculate the position vector $\overrightarrow{MA} = \overrightarrow{A} - \overrightarrow{M} = (1 - (-2))i + (2 - 4)j + (-3 - (-6))k = 3i - 2j + 3k$.Now,calculate the cross product $\overrightarrow{MA} 	imes \overrightarrow{F} = \begin{vmatrix} i & j & k \ 3 & -2 & 3 \ 2 & -6 & 5 \end{vmatrix}$.$= i((-2)(5) - (3)(-6)) - j((3)(5) - (3)(2)) + k((3)(-6) - (-2)(2))$.$= i(-10 + 18) - j(15 - 6) + k(-18 + 4)$.$= 8i - 9j - 14k$.

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

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