Part A.
Given:
P = (5, 4), Q = (7, 3), R = (8, 6), S = (4, 1)
Let's find the component of the vector PQ + 5RS.
To find the component of the vector, we have:
[tex]=\lparen Q_1-P_1,Q_2-P_2)=<7-5,3-4>[/tex]
For vector RS, we have:
[tex]=\lparen S_1-R_1,S_2-R_2)=<4-8,1-6>[/tex]
Hence, to find the vector PQ+5RS, we have:
[tex]\begin{gathered} =<7-5,3-4>+5<4-8,1-6> \\ \\ =\left(2,-1\right)+5\left(-4,-5\right) \\ \\ =\left(2,-1\right)+\left(5\ast-4,5\ast-5\right) \\ \\ =\left(2,-1\right)+\left(-20,-25\right) \end{gathered}[/tex]
Solving further:
[tex]\begin{gathered} =<2-20,-1-25> \\ \\ =<-18,-26> \end{gathered}[/tex]
Therefoee, the component of the vector PQ+5RS is:
<-18, -26>
• Part B.
Let's find the magnitude of the vector PQ+5RS.
To find the magnitude, apply the formula:
[tex]m=\sqrt{\left(x^2+y^2\right?}[/tex]
Thus, we have:
[tex]\begin{gathered} m=\sqrt{\left(-18^2+-26^2\right?} \\ \\ m=\sqrt{324+676} \\ \\ m=\sqrt{1000} \\ \\ m=\sqrt{10\ast10^2} \\ \\ m=10\sqrt{10} \end{gathered}[/tex]
Therefore, the magnitude of the vector is:
[tex]10\sqrt{10}[/tex]
ANSWER:
Part A. <-18, -26>
Part B. 10√10
