Four forces act on a stationary particle at t | vedclass

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(C) The resultant force $\overrightarrow{F}$ is the vector sum of all individual forces acting on the particle.$\overrightarrow{F} = \overrightarrow{F

Vedclass · Accepted Answer

$xz$-plane

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(C) The resultant force $\overrightarrow{F}$ is the vector sum of all individual forces acting on the particle.$\overrightarrow{F} = \overrightarrow{F_1} + \overrightarrow{F_2} + \overrightarrow{F_3} + \overrightarrow{F_4}$$\overrightarrow{F} = (3\hat{i} - \hat{j} + 9\hat{k}) + (2\hat{i} - 2\hat{j} + 16\hat{k}) + (9\hat{i} + \hat{j} + 18\hat{k}) + (\hat{i} + 2\hat{j} - 18\hat{k})$Summing the components:$x$-component: $3 + 2 + 9 + 1 = 15$$y$-component: $-1 - 2 + 1 + 2 = 0$$z$-component: $9 + 16 + 18 - 18 = 25$Thus,$\overrightarrow{F} = 15\hat{i} + 0\hat{j} + 25\hat{k}$.Since the $y$-component is $0$,the particle will move in the plane where $y = 0$,which is the $xz$-plane.

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