## Related questions with answers

Use a graphing utility to find the magnitude and direction angles of the resultant of forces $\mathbf{F}_1$ and $\mathbf{F}_2$ with initial points at the origin. The magnitude and terminal point of each vector are given.

Vector | Magnitude | Terminal Point |
---|---|---|

$\mathbf{F}_1$ | $50 \mathrm{lb}$ | $(10,5,3)$ |

$\mathbf{F}_2$ | $80 \mathrm{lb}$ | $(12,7,-5)$ |

Solution

Verified(1) Let $M_1$ be the magnitude per unit vector for $F_1$.

$F_1 = M_1\left\langle 10, 5, 3 \right\rangle$

$M_1 = M* \frac{1}{\left\|F_1\right\|} = 50 lb* \frac{1}{\sqrt{10^2 + 5^2 + 3^2}}=50 lb * \frac{1}{\sqrt{134}} \approx 4.319 lb$ per 1 unit scalar.

Let $M_2$ be the magnitude per unit vector for $F_2$.

$F_2 = M_2\left\langle 12, 7, -5 \right\rangle$

$M_2 = M* \frac{1}{\left\|F_2\right\|} = 80 lb* \frac{1}{\sqrt{12^2 + 7^2 + 5^2}}=80 lb * \frac{1}{\sqrt{218}} \approx 5.418 lb$ per 1 unit scalar.

$F_1 = 4.319 \left\langle 10, 5, 3 \right\rangle = \left\langle 43.19, 21.59, 12.95\right\rangle$

$F_2 = M_2\left\langle 12, 7, -5 \right\rangle = 5.418 \left\langle 12, 7, -5 \right\rangle = \left\langle 65.02 ,37.93 ,- 27.09\right\rangle$

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