Step 1

We are given the value of force $F = 350\;{\rm{lb}}$.

We are asked to express the force F as a Cartesian vector.

 
Step 2

The location of point A from the above diagram is.

\[{r_A} = \left( {0i + 0j + 35k} \right)\;{\rm{ft}}\]

The location of point B from the above diagram is.

\[{r_B} = \left( {50\sin \theta i + 50\cos \theta j + 0k} \right)\;{\rm{ft}}\]

Plugging the known values in the above expression.

\begin{array}{l} {r_B} = \left( {50\sin \left( {20^\circ } \right)i + 50\cos \left( {20^\circ } \right)j + 0k} \right)\;{\rm{ft}}\\ {r_B} = \left( {17.1i + 47j + 0k} \right)\;{\rm{ft}} \end{array}

The position vector among points A and B is calculated as,

\begin{array}{l} {r_{AB}} = {r_B} - {r_A}\\ {r_{AB}} = \left( {17.1i + 47j + 0k} \right)\;{\rm{ft}} - \left( {0i + 0j + 35k} \right)\;{\rm{ft}}\\ {r_{AB}} = \left( {\left( {17.1 - 0} \right)i + \left( {47 - 0} \right)j + \left( {0 - 35} \right)k} \right)\;{\rm{ft}}\\ {r_{AB}} = \left( {\left( {17.1} \right)i + \left( {47} \right)j - \left( {35} \right)k} \right)\;{\rm{ft}} \end{array}
 
Step 3

To find the magnitude of position vector we will use the following relation.

\[r = \sqrt {{r_x}^2 + {r_y}^2 + {r_z}^2} \]

Plugging in the known values in the above relation.

\begin{array}{l} r = \sqrt {{{\left( {17.1\;{\rm{ft}}} \right)}^2} + {{\left( {47\;{\rm{ft}}} \right)}^2} + {{\left( { - 35\;{\rm{ft}}} \right)}^2}} \\ r = 61.04\;{\rm{ft}} \end{array}

To find the unit vector we will use the following relation.

\[u = \left( {\frac{{{r_x}}}{r} + \frac{{{r_y}}}{r} + \frac{{{r_z}}}{r}} \right)\]

Plugging in the known values in the above relation.

\begin{array}{l} u = \left( {\frac{{17.1}}{{61.04}}i + \frac{{47}}{{61.04}}j + \frac{{35}}{{61.04}}k} \right)\\ u = \left( {0.28i + 0.76j + 0.57k} \right) \end{array}
 
Step 4

To find the Cartesian vector of force we will use the following relation.

\[{F_{BA}} = F \cdot u\]

Plugging in the known values in the above relation.

\[\begin{array}{l} {F_{BA}} = \left( {350\;{\rm{lb}}} \right) \cdot \left( {0.28i + 0.76j - 0.57k} \right)\\ {F_{BA}} = \left( {98i + 266j - 199.5k} \right)\;{\rm{lb}} \end{array}\]