How Do You Calculate Vector Sums and Differences Graphically?

[arildno]

Note that when you draw (X+Y) and (X-Y) in this manner, you draw up the diagonals of the parallellogram (in this case, a rectangle!) given by X and Y.

 

[arildno]

So, just to finish this:
The drawn line segment is the "translated (3,4)", where by (3,4) I mean the vector with its foot in the origin, and its arrowhead in (3,4).
(You've translated it by dragging it down along Y)

 

[soul814]

I'm confused by what you said. The resultant (R) of Vector A and Vector B is a straight line from the starting point to the ending point. If its two points its usually a diagonal.

 

[soul814]

Ohh a negative would flip the direction of the Y axis right and since vectors are allowed to move anywhere as long as its direction and magnitude is the same I can move it upward to (0,0) and (0,4) then shift it over to (3,0) and (3,4)

therefore the vector will be starting at (0,0) to (3,4)

right? the magnitude would still be 5 and the angle would still be 53 but it would be positive this time

 

[arildno]

Now look:
1. If you add Y and (X-Y) together, then your resultant is X, the diagonal in a parallellogram determined by Y and (X-Y).
(This is what I asked you to draw)
(X-Y) itself is the vector stretching from the origin up to (3,4)
(The leg adjoining Y in the origin in the parallellogram determined by them.)
2. You could also look at X-Y as X+(-Y).
Adding the vector (-Y) to X is what you did in your next to previous post; your resulting diagonal (in the parallellogram determined by X and (-Y)) is then (X-Y).

 

[soul814]

ehh did a seach on google, since at first I didnt get what you said, now after looking at some pictures I think I get what you mean.

First picture (A + B)
|--->
|
|
\/

Second Picture (A + (-B))
/\
|
|
|
|--->

Third Picture Shift (A) Upwards
/\--->
|
|
|
|

Now Draw the Line
/\--->
|.../
|.../
|../
|/

 

[arildno]

Precisely!

 

[soul814]

so magnitude and direction are still the same though right?

except its positive degrees

 

[arildno]

Yes, this is because X and Y are at right angles to each other.
(It is not true in general)

 

[arildno]

Note:
The "direction" is not the "same" since it uses positive degrees rather than negative degrees.

 

[soul814]

actually the angles different since

$$tan = opp/adj$$

so its $$tan^-^1 = 3/4$$

 

[arildno]

Nope, the angle between the vector (3,4) and the x-axis has "4" as the opposite side, and "3" at the adjecent side

 

[soul814]

(3,4) is (x,y)

so X-Axis is 3
and Y-Axis is 4

 

[arildno]

Definitely!
But its angle with respect to the x-axis, is to regard the triangle with hypotenuse up to (3,4), the vertical line segment down to (3,0) (and then along the horizontal to the origin).
What you're looking at is the vector's angle with respect to the y-axis, not with respect to the x-axis.

 

[soul814]

Ahh I see you calculate this

/\--->
|.../
|../
|./
|/X

and to calculate that you would have to shift the vectors
.../\
.../.|
../..|
./...|
/X..|

therefore its still $$tan = 4/3$$

 

[arildno]

Then we agree?

 

[soul814]

yep lol now there's this odd problem that I've been trying to solve but I can't get it. The answers in the back of the book but there's no explanation.

Each of the displacement vectors A and B shown in Figure P3.3 has a magnitude of 3. Graphically find (a) A + B (b) A - B (c) blah blah (d) blah blah

I think If I got one of it I would get the rest.

A is the diagonal line
B is Bolded the vertical line

/\
|.../
|.../
|.../
|../
|./
|/30(degrees)____

 

[arildno]

Well, it's just the same procedures really; are you supposed to answer with a diagram or with coordinate values?

 

[soul814]

this one isn't hw, i just want to get a better understanding of how to do it. the answers in numbers and there's a degree angle too.

I know how to move it but then you can't do the theorem with this.

...|
...|
^...|
|.../
|.../
|../
|/
----------- >

 

[arildno]

True enough, but let's find the coordinates of the skew line (A).
We know that its length is 3, and the angle to the x-axis is 30.
This means that its coordinates is:
$$(3\cos(30),3\sin(30))$$
You are now able to find the coordinates to for example, the sum of A+B

 

[soul814]

umm i did that... i get a different answer from the book

(2.6,1.5)

the other one is three but the same thing can be done to found it
3cos(90),3sin(90)

(0,3)

 

[arildno]

(2.6,1.5) looks right;
what does your calculator say when you type in cos(30)?
Multiply that number with 3.

 

[soul814]

i get .8660254038

*3 = 2.598076211

How would you get the answer after that?

Rx = Ax + Bx = 2.6 + 0
Ry = Ay + By = 1.5 + 3

Now I add Rx + Ry = 7.1

Answer in book is 5.2m at 60 above x-axis

 

[arildno]

HAVE YOU FORGOTTEN PYTHAGORAS?

The resultant vector is:
$$(\frac{3\sqrt{3}}{2},\frac{9}{2})$$
The length is therefore:
$$\sqrt{\frac{27}{4}+\frac{81}{4}}=\frac{\sqrt{108}}{2}$$

 

[soul814]

how did u get the resultant vector?

 

[arildno]

$$\cos(30)=\frac{\sqrt{3}}{2},\sin(30)=\frac{1}{2}$$
(This is a rather well-known relation; you'll it later on)
Hence A has coordinates $$(\frac{3\sqrt{3}}{2},\frac{3}{2})$$
Summing A with B (coord. (0,3)) yields the resultant vector).

 

[soul814]

ahh indeed it does.. this makes sense now :D

yes I learned cos(30) = 3/2 in precalculus the 30-60-90 triangle

 

[soul814]

wohoo got B) correct how would you get the degree though?

 

[arildno]

How do you think?

 

[soul814]

ummm I know all the measurements of the sides yet no angles except that 30 degrees, but its on the outside