**Vector A⃗ is 2.80 cm long and is 60.0∘ above the x-axis in the first quadrant….**

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## Question “Vector A⃗ is 2.80 cm long and is 60.0∘ above the x-axis in the first quadrant….”

Vector A⃗ is 2.80 cm long and is 60.0∘ above the x-axis in the

first quadrant. Vector B⃗ is 1.90 cm long and is 60.0∘ below the

x-axis in the fourth quadrant (the figure (Figure 1) Use components

to find the magnitude of B⃗ −A⃗

Use components to find the direction of B⃗ −A⃗ .

Sketch the vector subtraction *C*⃗ =*B*⃗

−*A*⃗ .

## Answer

Let’s say that Vector is R, and it makes angle with +x axis. Then its components are:

Rx = Horizontal component = R*cos

Ry = Vertical component =R*sin

Use the following rule:

Vector A = 2.80cm, 60.0 degrees above +x axis. (In the 1st quadrant both vector x and vector y are epositive).

Ax = A*cos = 2.80*cos 60 deg = 1.40 cm

Ay = A*sin = 2.80*sin 60 deg = 2.425 cm

Vector B = 1.90cm, 60.0 degrees below +x axis. (In the 4th quadrant, x is +ve while y is negative.)

Bx = B*cos = 1.90*cos 60 deg = 0.95 cm

By = -B*sin = -1.90*sin 60 deg = -1.645 cm

We now need to figure out

C = B – A = B + (-A).

A = Ax i + Ay j

(-A) = -Ax i – Ay j = -1.40 i – 2.425 j

B = Bx i + By j = 0.95 i – 1.645 j

So,

C = (Bx – Ax) i + (By – Ay) j

C = (0.95 – 1.40) i + (-1.645 – 2.425) j

C = -0.45 i – 4.07 j

The magnitude of vector C will be:

|C| = |B – A|

C

C

Direction of vector C is:

Direction = 180 + arctan (|Cy|/|Cx|)

Direction = 180 + arctan (4.07/0.45) = 180 + 83.7

**Direction = 263.7CCW starting at +ve x-axis**

Part C.

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## Conclusion

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