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

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

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 ?

To find the magnitude and direction of the vector PQ−→−, we can use the following steps:

Step 1: Determine the coordinates of the vector PQ−→− by subtracting the coordinates of point P from the coordinates of point Q. In this case, point P has coordinates (-6, 0) and point Q has coordinates (2, 4). Subtracting the corresponding coordinates gives us (2 - (-6), 4 - 0) = (8, 4).

Step 2: Calculate the magnitude of the vector PQ−→− using the formula: magnitude = √(x^2 + y^2), where x and y are the components of the vector. In this case, the components are 8 and 4, respectively. Plugging these values into the formula gives us: magnitude = √(8^2 + 4^2) = √(64 + 16) = √80 ≈ 8.9 units.

Step 3: Calculate the direction of the vector PQ−→− using the formula: direction = arctan(y / x), where y and x are the components of the vector. In this case, the components are 8 and 4, respectively. Plugging these values into the formula gives us: direction = arctan(4 / 8) ≈ 26.6° north of east.

Therefore, the magnitude and direction of the vector PQ−→− are approximately 8.9 units and 26.6° north of east.
Hence, the correct option is a. 8.9 units, 26.6° north of east.