What is the magnitude and direction of the PQ−→− with tail and head points P(-6, 0) and Q(2, 4)?
Select one:
a. 8.9 units, 26.6° north of east
b. 8.9 units, 63.4° north of east
c. 13.4 units, 0° east
3 answers
Nevermind it is A
right 8 and up 4
magnitude = sqrt (64+16)= sqrt (4^2* 4 + 4^2) = 4 sqrt 5= 8.9
tan angle n of e = 1/2
so angle n of e = 26.6
magnitude = sqrt (64+16)= sqrt (4^2* 4 + 4^2) = 4 sqrt 5= 8.9
tan angle n of e = 1/2
so angle n of e = 26.6
vector PQ = (2-(-6), 4-0) = (8,4)
magnitude = √(64+16) = √80 = appr 8.9 , ruling out c.
angle = tan^-1 (4/8) = 26.57° from the x-axis
what angle description describes that ?
magnitude = √(64+16) = √80 = appr 8.9 , ruling out c.
angle = tan^-1 (4/8) = 26.57° from the x-axis
what angle description describes that ?
3 answers