Question

You are given a vector in the xy plane that has a magnitude of 89.0 units and a y component of -60.0 units.

What are the two possibilities for its x component?

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.

Specify the direction of the vector.

Answer 1

Let, given vector = R = Rx i + Ry j

given, |R| = 89.0 units

Ry = -60.0 units

then, |R| = sqrt(Rx^2 + Ry^2)

89.0^2 = Rx^2 + (-60.0)^2

Rx = sqrt(89.0^2 - 60.0^2)

Rx = 65.7 units

So, the two possibilities for its x component are: 65.7 , -65.7

Now, given the x component is known to be positive.

then, Rx = 65.7 units

Let, another vector will be:

R' = R'x i + R'y j

Given, R + R' = 80.0 units long and points entirely in the −x direction

R + R' = -80.0 i + 0 j

using known values:

(65.7 i - 60 j) + (R'x i + R'y j) = -80.0 i + 0 j

(65.7 + R'x) i + (R'y - 60) j = -80.0 i + 0 j

By comparing x-component on both sides,

65.7 + R'x = -80.0

R'x = -80.0 - 65.7

R'x = -145.7 units

By comparing y-component on both sides,

-60 + R'y = 0

R'y = 60.0 units

So, R' = -145.7 i + 60 j

Magnitude of vector will be:

|R'| = sqrt((-145.7)^2 + 60^2) = 157.57

In three significant figure:

|R'| = 158 units

Direction of vector will be:

= arctan(R'y/R'x) = arctan(60/145.7)

= 22.4 deg above -x axis

= 157.6 deg above +x axis in counter-clock wise direction.

"Let me know if you have any query."