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Determine the angle $\theta$ between the force and the line $A B .$

$\vec{r}_{A B}=\vec{r}_{B} - \vec{r}_{A}$

$=(0\hat i+0\hat j+4 k)-(0\hat i+3\hat j+0 k)$

$=(0 \hat{\imath}-3 \hat{\jmath}+4 K)$

$\left|r_{A B}\right|=5 m$

$\vec{r}_{A c}=\vec{r}_{c}-\vec{r}_{A}$

$=(4 \hat{\imath}+0 \hat{\jmath}+0 k)-(0 i+3 j+0 k)$

$=(4\hat i-3\hat j+0 k) m$

$\left|r_{A C}\right|=5 \mathrm{m}$

$\vec{r}_{A B} \cdot \vec{r}_{A C} =\left|r_{A B}\right|\left|r_{A C}\right| \cos \theta \rightarrow \left(0\hat i -3 \hat j+4 k\right) \cdot(4 \hat i-3\hat j+0 k)=5(5) \cos \theta$

$∴ {\theta}=68 \cdot 9^{\circ}$

Two cables exert forces on the pipe. Determine the magnitude of the projected component of $\mathbf{F}_{1}$ along the line of action of $\mathbf{F}_{2}$ .

$\vec{F}=\left(30 \cos 30^{\circ} \sin 30^{\circ} i\right)+\left(30 \cos 30 \cos 30\hat j\right)$

$-(30 \sin 30 k)$

$=(13 \hat{\imath}+22.5 \hat j-15 k) N$

* unit vector of $F_{2}$:-

$\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma =1 \quad \longrightarrow\alpha=45^{\circ}$ or $\alpha=135^{\circ}$

$\overrightarrow{U} \ {F_{2}}=\cos 135 \hat i+\cos 60^{\circ}\hat j+\cos 60^{\circ} k$

$=(-0.707\hat i)+0.5\hat j+0.6 k )$

$F_∥=F_{1} \cdot \overrightarrow{U} F_{2} = (13\hat i+22 \cdot 5\hat j-15 k) \cdot(-0.707\hat i+0.5\hat j+0.6 k)$

$=-5.44 N$

$\left|F_∥\right|=5.44 N$