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I need some help on solving a simple vector force physics question.
Given a force vector (x, y), how to calculate the magnitude of the force apply on the direction (i, j) ?
(For example, given (1, 0) the magnitude of the force apply on the direction (0, 1) should be 0, and given (1, 0) the magnitude of the force apply on the direction (-1, 0) should be -1)
Thank you very much!
It is just vector projection on another vector. It could be calculated as
ProjLength = DotProduct(F, A) / Length(A)
where F is force vector, A is direction vector
As far as I remember, in middle school we were taught to use cosine of angle between force and moving directions.
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This is a follow-up question to this question.
Taking the following image as an example:
What I know:
x and y coordinates of points D, E, and P.
Therefore, I also know slope and intercept of D-E line
What I want to know:
x and y coordinates of point Q. (This is the point which crosses the D-E line).
Notation P=[px,py], D=[dx,dy], E=[ex,ey], Q=[qx,qy]
First:
R=P-D=[px-dx, py-dy]=[rx,ry]
K=E-D=[ex-dx, ey-dy]=[kx, ky]
Then
z=dot(R,K)/dot(K,K)=(rx*kx+ry*ky) / (kx*kx+ky*ky)
Finally
Q=D+z*K=[dx+z*kx, dy+z*ky]
The R is vector which start on point D and ends on point K, the K is vector which start on point D and ends on point E. Using this we made scalar projection to calculate result Q. More info about concept here
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Please forgive my lack of knowledge, which i think it's one of those basic formula related to Trigonometry.
Let's look at visual example:
I have 5 lines, with their line equation (let's say they have zero offset ok)
how can i calculate the Theta of each line equation make (in Pi)?
also I have seen this:
Are they generated from Theta of line equations? or it's another theory which help to find the theta?
much appreciate your time and effort
For equation
y = k * x
tg(Theta) = k
and
Theta = Arctg(k) //arctangent function
General line equation
A * x + B * y + C = 0
(It is more general than y=ax+b and includes cases of vertical and horizontal lines)
Theta = atan2(A, B)
(function atan2 or ArcTan2 exists in math libraries of many programming languages)
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i'm trying to practice my math for my upcoming exam, and i have an example that i'm trying to work out.
I'm trying to find an angle of exactly 90% (making vector B perpendicular to vector A)
A = (1, 3, 2)
B = (2, x, -2)
I've tried x = 0, x = 0.5, x = 1. I can't seem to get the Dot product to equal 0 (making them perpendicular of course).
Can anyone shed some light on how i can find "x" to make A and B 90 degree between each other?
Thanks!
This really should be on math.stackexchange, but someone else can move it. Just compute the dot product of the vectors:
(1, 3, 2) dot (2, x, -2) = 0
2 + 3x - 4 = 0
x = 2/3
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Given (x,y) that is (3,4) , velocity is 1 m/sec and given angle is 15 degree. what will be next coordinate (x', y') after 1 sec ?
Anyone please help!
Assuming the angle is measure w.r.t. the x-axis, then:
the total distance is v*t = 1m
displacement in x-direction: cos(15)*1m=.97m
displacement in y-direction: sin(15)*1m=.25m
So the new location is (3.97, 4.25).
#Alexander-Vogt's answer is close but I think it is missing a conversion from angles to radians. Here is matlab code that provides the answer:
pi=3.1415926535897932384626433832795028841971;
x=3;y=4;
t=1; % time
s=1*t; % speed * time
a=15; % angle of movement
a=a/360 * pi*2; % convert to radians
x=cos(a)*s+x;
y=sin(a)*s+y;
fprintf("%f %f\n",x,y);
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I want to rotate a plane represented by the equation z = 6 , by n degrees along y axis and find the new equation of the plane. how can this be done?
Thanks
Base point (0,0,6) after rotation will lie in XZ plane with coordinates
(x0, y0, z0) = (-6*sin(Fi), 0, 6*cos(Fi))
normal vector
n = (A,B,C) = (-sin(Fi), 0, cos(Fi))
so new plane equation is (for explanation see beginning of the article)
A*(x-x0)+B*(y-y0)+C*(z-z0)=0
or
-sin(Fi)*x + cos(Fi)*z - 6 = 0