Three forces i + 2j - 3k,2i + 3j + 4k,and i - | vedclass

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(B) Let the forces be $\vec{F_1} = i + 2j - 3k$,$\vec{F_2} = 2i + 3j + 4k$,and $\vec{F_3} = i - j + k$.The resultant force $\vec{F} = \vec{F_1} + \vec

Vedclass · Accepted Answer

$6\sqrt{10}$

Vedclass · Answer

(B) Let the forces be $\vec{F_1} = i + 2j - 3k$,$\vec{F_2} = 2i + 3j + 4k$,and $\vec{F_3} = i - j + k$.The resultant force $\vec{F} = \vec{F_1} + \vec{F_2} + \vec{F_3} = (1+2+1)i + (2+3-1)j + (-3+4+1)k = 4i + 4j + 2k$.The particle is at point $A(0, 1, 2)$ and we need the moment about point $B(1, -2, 0)$.The position vector $\vec{r}$ of the point of application relative to the point about which the moment is calculated is $\vec{r} = \vec{OA} - \vec{OB} = (0-1)i + (1-(-2))j + (2-0)k = -i + 3j + 2k$.The moment $\vec{M} = \vec{r} 	imes \vec{F} = \begin{vmatrix} i & j & k \ -1 & 3 & 2 \ 4 & 4 & 2 \end{vmatrix}$.Calculating the determinant: $\vec{M} = i(6 - 8) - j(-2 - 8) + k(-4 - 12) = -2i + 10j - 16k$.The magnitude of the moment is $|\vec{M}| = \sqrt{(-2)^2 + 10^2 + (-16)^2} = \sqrt{4 + 100 + 256} = \sqrt{360} = \sqrt{36 	imes 10} = 6\sqrt{10}$.

Three forces $i + 2j - 3k$,$2i + 3j + 4k$,and $i - j + k$ are acting on a particle at the point $(0, 1, 2)$. The magnitude of the moment of the forces about the point $(1, -2, 0)$ is

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