Question

Part A
Given that P=(5,4), Q=(7,3), R=(8,6), and S=(4,1), find the component form of the vector PQ+3RS.
a. <-10,-16>****
b. <-6,4>
c. <-2,-6>
d. <14,14>

Part B
Use the information from Part A to find the magnitude of the vector PQ+3RS.
a. 356
b. sqrt(26)
c. 2*sqrt(10)
d. 2*sqrt(89)

I know that Part A is correct, but I do not understand how to find the magnitude.

Answer 1

I would also appreciate if someone could better explain how to get Part A because my teacher did not explain this very well.

Answer 2

In general if you have points P(a,b) and Q(c,d)
then vector PQ = <c-a , d-b>
Notice I went from P to Q, so the subtraction is "destination x - starting x" , same for the y
Then to get vector QP we would get <a-c, b-d>

for yours, PQ = <7-5, 3-4> = <2,-1>

to find the magnitude of vector <x,y> find √( x^2 + y^2)
e.g. |PQ| = √(2^2 + (-1)^2 ) = √5

since your part a) is correct simply follow the above rule, I can see the answer

Answer 3

a. PQ=(7-5)x+(3-4(y)=2 ,-1
RS=(4-8)x+(1-6)y=-4,-5
PQ+3RS=2 ,-1 +3( -4,-5)=(2-14,-1-15)=10,-16

magnitude= sqrt(10^2+(-16)^2)=sqrt(356)
factoring 356
2*178
2*2*89
magnitude=2sqrt(89)