Answer
$\mathrm{F}_p=10.5 \mathrm{lb}$
Work Step by Step
$\begin{aligned} \vec{\gamma}_{B C} & =\{6 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}\} \mathrm{ft} \\ \overrightarrow{\mathrm{F}} & =100 \frac{\{-6 \hat{\mathrm{i}}+8 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}\}}{\sqrt{(-6)^2+8^2+2^2}} \\ & =\{-58.83 \hat{\mathrm{i}}+78.45 \hat{\mathrm{j}}+19.61 \hat{\mathrm{k}}\} \mathrm{lb} \\ \mathrm{F}_P & =\overrightarrow{\mathrm{F}} \cdot \vec{\mu}_{B C}=\overrightarrow{\mathrm{F}} \cdot \frac{\vec{\gamma}_{B C}}{\left|\vec{\gamma}_{B C}\right|}=\frac{-78.45}{7.483}=-10.48 \\ \mathrm{~F}_P & =10.5 \mathrm{lb}\end{aligned}$
