You are given a vector in the xy plane that has a magnitude of 90.0 units and a y component of -61.0 units. A) Assuming the x component is known to be positive, specify the magnitude of the vector which, if you add it to the original one, would give a resultant vector that is 80.0 units long and points entirely in the −x direction. B)Specify the direction of the vector. Express your answer using three significant figures.

A) magnitude = sqrrt(147.1^2 + 61^2) = 159.2 units

B)22.5° (clockwise form -ve X axis)

Explanation:

given vector = V1 = x i - 60 j

magnitude of V1 = 90

x - component can be found out by resultant formula

90^2 = x^2 + (-60)^2

x = 67.08 = 67.1 units (3sf)

FOR THE VECTOR 80 UNITS IN -VE X DIRECTION

The X component is -80(1)

The Y component is 0 (2)

For the x- component of new added vector:

(1) x + 67.1 = -80

x = -147.1 = -147.1

For the y- component of new added vector:

(2) y - 61  = 0

y = 61.0 (3sf)

the new added vector is  = -147.1 i + 61 j

magnitude = sqrrt(147.1^2 + 61^2) = 159.2 units

direction = arctan (61 / 147.1)

               = 22.5° (clockwise form -ve X axis)

are there options? cause you ain’t. real specific

double the distance = divide strength by 4