If a particle is acted upon by forces of magn | vedclass

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(NONE) The resultant force $\vec{F}$ is the sum of the two force vectors.
First,find the unit vectors in the given directions:
$\hat{u}_1 = \frac{-\

Vedclass · Accepted Answer

$-8$

Vedclass · Answer

(NONE) The resultant force $\vec{F}$ is the sum of the two force vectors.
First,find the unit vectors in the given directions:
$\hat{u}_1 = \frac{-\hat{i} - 2\hat{j} + 2\hat{k}}{\sqrt{(-1)^2 + (-2)^2 + 2^2}} = \frac{-\hat{i} - 2\hat{j} + 2\hat{k}}{3}$
$\hat{u}_2 = \frac{2\hat{i} - 3\hat{j} - 6\hat{k}}{\sqrt{2^2 + (-3)^2 + (-6)^2}} = \frac{2\hat{i} - 3\hat{j} - 6\hat{k}}{7}$
Now,calculate the force vectors:
$\vec{F}_1 = 6 \hat{u}_1 = 2(-\hat{i} - 2\hat{j} + 2\hat{k}) = -2\hat{i} - 4\hat{j} + 4\hat{k}$
$\vec{F}_2 = 7 \hat{u}_2 = 1(2\hat{i} - 3\hat{j} - 6\hat{k}) = 2\hat{i} - 3\hat{j} - 6\hat{k}$
Resultant force $\vec{F} = \vec{F}_1 + \vec{F}_2 = (-2+2)\hat{i} + (-4-3)\hat{j} + (4-6)\hat{k} = 0\hat{i} - 7\hat{j} - 2\hat{k}$
Displacement vector $\vec{d} = \vec{PQ} = (5-2)\hat{i} + (-1 - (-1))\hat{j} + (1 - (-3))\hat{k} = 3\hat{i} + 0\hat{j} + 4\hat{k}$
Work done $W = \vec{F} \cdot \vec{d} = (0\hat{i} - 7\hat{j} - 2\hat{k}) \cdot (3\hat{i} + 0\hat{j} + 4\hat{k}) = (0)(3) + (-7)(0) + (-2)(4) = -8$ units.

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