SOLUTION
The coordinate of the vector P and Q are
[tex]\begin{gathered} P(-13,\text{ 11)} \\ \text{And } \\ Q(-18,2) \end{gathered}[/tex]To find the vector PQ. we have
[tex]\begin{gathered} t=\bar{PQ} \\ PQ\text{ is having the coordinate } \\ PQ=(-18-(-13),2-11)=(-5,-9) \end{gathered}[/tex]To find the vector, we use
[tex]\begin{gathered} r=\sqrt[]{x^2+y^2} \\ \text{Where } \\ x=-5,y=-9 \\ r=\sqrt[]{(-5)^2+(-9)^2}=\sqrt[]{25+81}=\sqrt[]{106}=10.296 \end{gathered}[/tex]Then we obtain the angle using
[tex]\begin{gathered} \text{tan}\theta=(\frac{y}{x})_{} \\ \text{Substituting the value of x and y, we have } \\ \tan \theta=(\frac{9}{5})=\tan \theta=(1.8) \end{gathered}[/tex]Hence
[tex]\begin{gathered} \tan \theta=1.8 \\ \theta=\tan ^{-1}(1.8) \\ \theta=60.945 \end{gathered}[/tex]Hence
The vector in trigonometry form will be
[tex]\begin{gathered} t=r(i\cos \theta+j\sin \theta) \\ \text{Then} \\ t=10.296\cos 60.945i+10.296\sin 60.945j \end{gathered}[/tex]Therefore
t= 10.296 cos 60.945 i + 10.296 sin 60.945j
Answer: Option C(third option ).
