If overrightarrow{F_1} = i - j + k, overright | vedclass

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(C) Step $1$: Calculate the sum of the vectors $\overrightarrow{F_1}, \overrightarrow{F_2},$ and $\overrightarrow{F_3}$.$\Sigma \vec{F} = \overrightar

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(C) Step $1$: Calculate the sum of the vectors $\overrightarrow{F_1}, \overrightarrow{F_2},$ and $\overrightarrow{F_3}$.$\Sigma \vec{F} = \overrightarrow{F_1} + \overrightarrow{F_2} + \overrightarrow{F_3} = (i - j + k) + (-i + 2j - k) + (j - k) = (1-1)i + (-1+2+1)j + (1-1-1)k = 0i + 2j - k = 2j - k.$Step $2$: Calculate the vector $\overrightarrow{AB}$.$\overrightarrow{AB} = \vec{B} - \vec{A} = (6i + j - 3k) - (4i - 3j - 2k) = (6-4)i + (1 - (-3))j + (-3 - (-2))k = 2i + 4j - k.$Step $3$: Calculate the scalar (dot) product of $\Sigma \vec{F}$ and $\overrightarrow{AB}$.$(\Sigma \vec{F}) \cdot \overrightarrow{AB} = (2j - k) \cdot (2i + 4j - k) = (0)(2) + (2)(4) + (-1)(-1) = 0 + 8 + 1 = 9.$

If $\overrightarrow{F_1} = i - j + k,$ $\overrightarrow{F_2} = -i + 2j - k,$ $\overrightarrow{F_3} = j - k,$ $\vec{A} = 4i - 3j - 2k$ and $\vec{B} = 6i + j - 3k,$ then the scalar product of $(\overrightarrow{F_1} + \overrightarrow{F_2} + \overrightarrow{F_3})$ and $\overrightarrow{AB}$ will be:

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