Let the points A, B and P be (-2, 2, 4), (2, | vedclass

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(B) Given points are $A(-2, 2, 4)$,$B(2, 6, 3)$,and $P(1, 2, 1)$.The force vector $\overrightarrow{F} = \overrightarrow{AB} = (2 - (-2))\hat{i} + (6 -

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$3\sqrt{41}$

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(B) Given points are $A(-2, 2, 4)$,$B(2, 6, 3)$,and $P(1, 2, 1)$.The force vector $\overrightarrow{F} = \overrightarrow{AB} = (2 - (-2))\hat{i} + (6 - 2)\hat{j} + (3 - 4)\hat{k} = 4\hat{i} + 4\hat{j} - \hat{k}$.The position vector of $A$ with respect to $P$ is $\overrightarrow{r} = \overrightarrow{PA} = (-2 - 1)\hat{i} + (2 - 2)\hat{j} + (4 - 1)\hat{k} = -3\hat{i} + 0\hat{j} + 3\hat{k}$.The moment of the force about $P$ is $\vec{M} = \overrightarrow{PA} 	imes \overrightarrow{F}$.$\vec{M} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -3 & 0 & 3 \ 4 & 4 & -1 \end{vmatrix} = \hat{i}(0 - 12) - \hat{j}(3 - 12) + \hat{k}(-12 - 0) = -12\hat{i} + 9\hat{j} - 12\hat{k}$.The magnitude of the moment is $|\vec{M}| = \sqrt{(-12)^2 + 9^2 + (-12)^2} = \sqrt{144 + 81 + 144} = \sqrt{369} = \sqrt{9 	imes 41} = 3\sqrt{41}$.

Let the points $A, B$ and $P$ be $(-2, 2, 4), (2, 6, 3)$ and $(1, 2, 1)$ respectively. The magnitude of the moment of the force represented by $\overrightarrow{AB}$ and acting at $A$ about $P$ is

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