Calculating Vector Magnitude and Angle

[viet1919]

Homework Statement

What is the magnitude and angle to
Vector B - 2(Vector A)?

Homework Equations

A sub x: 44cos28 = 38.85
A sub y: 44sin28 = 20.66
B sub x: 26.5cos56 = 14.82 <-- (Does the negative sign matter in 26.5? My teacher said it's not)
B sub y: 26.5sin56 = 21.97

The Attempt at a Solution

I multiplied the Vector A's by two.
A sub x: 38.85 x 2 = 77.7
A sub y: 20.66 x 2 = 41.2

I then solved for the x and y components.
x: (77.7) + (14.82) = 92.52
y: (41.2) - (21.97) = 19.23

Then I solved for the magnitude and angle.
Magnitude:
[itex]\sqrt{}[/itex](92.52)² + (19.23)²= 94.50

Angle: tan^-1 = 19.23/92.52 --> 11.74°

Is this correct? I have a weird feeling that it's wrong. I feel like I am missing something some wear. I really can't tell.

 

[ehild]

Homework Statement

What is the magnitude and angle to
Vector B - 2(Vector A)?

Homework Equations

A sub x: 44cos28 = 38.85
A sub y: 44sin28 = 20.66
B sub x: 26.5cos56 = 14.82 <-- (Does the negative sign matter in 26.5? My teacher said it's not)
B sub y: 26.5sin56 = 21.97

Correct so far...

The Attempt at a Solution

I multiplied the Vector A's by two.
A sub x: 38.85 x 2 = 77.7
A sub y: 20.66 x 2 = 41.2

I then solved for the x and y components.
x: (77.7) + (14.82) = 92.52
y: (41.2) - (21.97) = 19.23

You have to find the vector [itex]\vec{B}[/itex] - 2[itex]\vec{A}[/itex]; subtracting twice of [itex]\vec{A}[/itex] from [itex]\vec{B}[/itex]. Do the same with both components: Evaluate B_x-2A_x and B_y-2A_y. These will be the components of a new vector [itex]\vec{C}[/itex], to determine the magnitude and angle of it.

Then I solved for the magnitude and angle.
Magnitude:
[itex]\sqrt{92.52^2 + 19.23^2}[/itex]= 94.50

Angle: tan^-1 = 19.23/92.52 --> 11.74°

Is this correct? I have a weird feeling that it's wrong. I feel like I am missing something some wear. I really can't tell.

ehild

 

[viet1919]

Oh so you're saying what I should've done was
x: 14.82 - 77.7
y: 21.97 - 41.2
Is that so? And If you looked at the 26.5cos56. I thought it was -26.5cos56. But my teacher said the sign negative sign didn't matter or something. What does he mean?

 

[viet1919]

At first it was this. Then I think we got rid of the negatives and made it positive.
Therefore using the equations

2(a sub x) + b sub x
and
2(a sub y) - b sub y

 

[ehild]

Oh so you're saying what I should've done was
x: 14.82 - 77.7
y: 21.97 - 41.2

Yes, it will be all right.

Is that so? And If you looked at the 26.5cos56. I thought it was -26.5cos56. But my teacher said the sign negative sign didn't matter or something. What does he mean?

First you calculated the x and y components of the vectors A and B. Then you did the operations with the components, multiplying by two and subtracting 2A from B.

Vectors can be add and subtract, and multiplied by scalars. You do it with the components, as the components are simply numbers. Then the results will be the components of the new vector.

ehild

 

[viet1919]

Ah ok I get it now! Wow it amazes me how you can sort of "play" with the numbers and still get the same answer. Thank you so much ehild.