Question

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 C⃗ =B⃗ −A⃗ .

Answer 1

Suppose given that Vector is R and it makes angle $\theta$ with +x-axis, then it's components are given by:

Rx = Horizontal component = R*cos $\theta$

Ry = Vertical component = R*sin $\theta$

Using above rule:

Vector A = 2.80 cm, 60.0 deg above +x axis (In 1st quadrant both x and y ar epositive)

Ax = A*cos $\theta$ = 2.80*cos 60 deg = 1.40 cm

Ay = A*sin $\theta$ = 2.80*sin 60 deg = 2.425 cm

Vector B = 1.90 cm, 60.0 deg below +x axis (In 4th quadrant x is +ve and y is negative)

Bx = B*cos $\theta$ = 1.90*cos 60 deg = 0.95 cm

By = -B*sin $\theta$ = -1.90*sin 60 deg = -1.645 cm

Now we need to find,

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

Magnitude of vector C will be:

|C| = |B - A|

|C| = sqrt ((-0.45)^2 + (-4.07)^2)

|C| = 4.0948 cm = 4.10 cm

Direction of vector C will be:

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

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

Direction = 263.7 CCW from +ve x-axis

Part C.

263. さん

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