Find the vector(s) 𝒘⃗ in 3D that simultaneously satisfy the following conditions:
a) ‖𝒘⃗ ‖ = 𝟏𝟎
b) 𝒘⃗ is perpendicular to the vector 〈𝟑, −𝟏, 𝟎〉
c) 𝒘⃗ forms an angle of 𝝅/𝟑 with the vector 〈𝟎, 𝟎, 𝟏〉
Put w = (x,y,z).
Condition (b) says 3x-y = 0, i.e. y = 3x.
Condition (c) means z = 0⋅x + 0⋅y + 1⋅z = cos(π/3)‖w‖‖(0,0,1)‖ = 0.5⋅10 = 5.
Condition (a) implies x^2 + (3x)^2 + 25 = 100, i.e., 10x^2 = 75 or x = ±sqrt(7.5).
In sum w = (±sqrt(7.5), ±3⋅sqrt(7.5), 5).
Note that there are two solutions, one with + and another with -.
Related
I have this vector
>>> vec
tensor([[0.2677, 0.1158, 0.5954, 0.9210, 0.3622, 0.4081, 0.4477, 0.7930, 0.1161,
0.5111, 0.2010, 0.3680, 0.1162, 0.1563, 0.4478, 0.9732, 0.7962, 0.0873,
0.9793, 0.9382, 0.9468, 0.0851, 0.7601, 0.0322, 0.7553, 0.4025, 0.3627,
0.5706, 0.3015, 0.1344, 0.8343, 0.8187, 0.4287, 0.5785, 0.9527, 0.1632,
0.2890, 0.5411, 0.5319, 0.7163, 0.3166, 0.5717, 0.5018, 0.5368, 0.3321]])
using this vector I want to generate 15 vectors which have a cosine similarity greater than 80%.
How can I do this in pytorch?
I modified the answer here, adding an extra dimension and converting from numpy to torch.
def torch_cos_sim(v,cos_theta,n_vectors = 1,EXACT = True):
"""
EXACT - if True, all vectors will have exactly cos_theta similarity.
if False, all vectors will have >= cos_theta similarity
v - original vector (1D tensor)
cos_theta -cos similarity in range [-1,1]
"""
# unit vector in direction of v
u = v / torch.norm(v)
u = u.unsqueeze(0).repeat(n_vectors,1)
# random vector with elements in range [-1,1]
r = torch.rand([n_vectors,len(v)])*2 -1
# unit vector perpendicular to v and u
uperp = torch.stack([r[i] - (torch.dot(r[i],u[i]) * u[i]) for i in range(len(u))])
uperp = uperp/ (torch.norm(uperp,dim = 1).unsqueeze(1).repeat(1,v.shape[0]))
if not EXACT:
cos_theta = torch.rand(n_vectors)* (1-cos_theta) + cos_theta
cos_theta = cos_theta.unsqueeze(1).repeat(1,v.shape[0])
# w is the linear combination of u and uperp with coefficients costheta
# and sin(theta) = sqrt(1 - costheta**2), respectively:
w = cos_theta*u + torch.sqrt(1 - torch.tensor(cos_theta)**2)*uperp
return w
You can check the output with:
vec = torch.rand(54)
output = torch_cos_sim(vec,0.6,n_vectors = 15, EXACT = False)
# test cos similarity
for item in output:
print(torch.dot(vec,item)/(torch.norm(vec)*torch.norm(item)))
I have this triangle:
I'm trying to get the vertex value highlighted in the green circle in order to draw that red line. Is there any equation that I can use to extract that value?
The centroid vertex G = (x=5.5, y=1.5)
The other vertex B = (x=0, y=1)
and the last vertex C = (x=7, y=0)
Any help would be appreciated. I know it might be a 5th grade math but I can't think of a way to calculate this point.
If you throw away the majority of the triangle and just keep the vector B->G and the vector B->C then this problem shows itself to be a "vector projection" problem.
These are solved analytically using the dot product of the 2 vectors and are well documented elsewhere.
Took me 2 days to figure this out, you basically need to get the slopes for the base vector and the altitude vector (centroid), then solve this equation: y = m * x + b for both vectors (the base + altitude). Then you'll get 2 different equations that you need to use substitution to get the x first then apply that value to the 2nd equation to get the y. For more information watch this youtube tutorial:
https://www.youtube.com/watch?v=VuEbWkF5lcM
Here's the solution in PHP (pseudo) if anyone is interested:
//slope of base
$m1 = getSlope(baseVector);
//slope of altitude (invert and divide it by 1)
$m2 = 1/-$m1;
//points
$x1 = $baseVector->x;
$y1 = $baseVector->y;
//Centroid vertex
$x2 = $center['x'];
$y2 = $center['y'];
//altitude equation: y = m * x + b
//eq1: y1 = (m1 * x1) + b1 then find b1
$b1 = -($m1 * $x1) + $y1;
//equation: y = ($m1 * x) + $b1
//eq2: y2 = (m2 * x2) + b2 then find b2
$b2 = -($m2 * $x2) + $y2;
//equation: y = ($m2 * x) + $b2;
//substitute eq1 into eq2 and find x
//merge the equations (move the Xs to the left side and numbers on the right side)
$Xs = $m1 - $m2; //left side (number of Xs)
$Bs = $b2 - $b1; //right side
$x = $Bs / $Xs; //get x number
$y = ($m2 * $x) + $b2; //get y number
I want to create a polynomial with given coefficients. This seems very simple but what I have found till now did not appear to be the thing I desired.
For example in such an environment;
n = 11
K = GF(4,'a')
R = PolynomialRing(GF(4,'a'),"x")
x = R.gen()
a = K.gen()
v = [1,a,0,0,1,1,1,a,a,0,1]
Given a list/vector v of length n (I will set this n and v at the begining), I want to get the polynomial v(x) as v[i]*x^i.
(Actually after that I am going to build the quotient ring GF(4,'a')[x] /< x^n-v(x) > after getting this v(x) from above) then I will say;
S = R.quotient(x^n-v(x), 'y')
y = S.gen()
But I couldn't write it.
This is a frequently asked question in many places so it is better to leave it here as an answer although the answer I have is so simple:
I just wrote R(v) and it gave me the polynomial:
sage
n = 11
K = GF(4,'a')
R = PolynomialRing(GF(4,'a'),"x")
x = R.gen()
a = K.gen()
v = [1,a,0,0,1,1,1,a,a,0,1]
R(v)
x^10 + a*x^8 + a*x^7 + x^6 + x^5 + x^4 + a*x + 1
Basically (that is, ignoring the specifics of your polynomial ring) you have a list/vector v of length n and you require a polynomial which is the sum of all v[i]*x^i. Note that this sum equals the matrix product V.X where V is a one row matrix (essentially equal to the vector v) and X is a column matrix consisting of powers of x. In Maxima you could write
v: [1,a,0,0,1,1,1,a,a,0,1]$
n: length(v)$
V: matrix(v)$
X: genmatrix(lambda([i,j], x^(i-1)), n, 1)$
V.X;
The output is
x^10+ax^8+ax^7+x^6+x^5+x^4+a*x+1
I am trying to create a binary tree from a lot of segments in 3d space sharing the same origin.
When merging two segments I want to have a specific angle between the lines to the child nodes.
The following image illustrates my problem. C shows the position of the parent node and A and B the child positions. N is the average vector of the vectors from C to A and C to B.
With a given angle, how can I determine point P?
Thanks for any help
P = C + t * ((A + B)/2 - C) t is unknown parameter
PA = A - P PA vector
PB = B - P PB vector
Tan(Fi) = (PA x PB) / (PA * PB) (cross product in the nominator, scalar product in the denominator)
Tan(Fi) * (PA.x*PB.x + PA.y*PB.y) = (PA.x*PB.y - PA.y*PB.x)
this is quadratic equation for t, after solving we will get two (for non-degenerate cases) possible positions of P point (the second one lies at other side of AB line)
Addition:
Let's ax = A.x - A point X-coordinate and so on,
abcx = (ax+bx)/2-cx, abcy = (ay+by)/2-cy
pax = ax-cx - t*abcx, pay = ay-cy - t*abcy
pbx = bx-cx - t*abcx, pby = by-cy - t*abcy
ff = Tan(Fi) , then
ff*(pax*pbx+pay*pby)-pax*pby+pay*pbx=0
ff*((ax-cx - t*abcx)*(bx-cx - t*abcx)+(ay-cy - t*abcy)*(by-cy - t*abcy)) -
- (ax-cx - t*abcx)*(by-cy - t*abcy) + (ay-cy - t*abcy)*(bx-cx - t*abcx) =
t^2 *(ff*(abcx^2+abcy^2)) +
t * (-2*ff*(abcx^2+abcy^2) + abcx*(by-ay) + abcy*(ax-bx) ) +
(ff*((ax-cx)*(bx-cx)+(ay-cy)*(by-cy)) - (ax-cx)*(by-cy)+(bx-cx)*(ay-cy)) =0
This is quadratic equation AA*t^2 + BB*t + CC = 0 with coefficients
AA = ff*(abcx^2+abcy^2)
BB = -2*ff*(abcx^2+abcy^2) + abcx*(by-ay) + abcy*(ax-bx)
CC = ff*((ax-cx)*(bx-cx)+(ay-cy)*(by-cy)) - (ax-cx)*(by-cy)+(bx-cx)*(ay-cy)
P.S. My answer is for 2d-case!
For 3d: It is probably simpler to use scalar product only (with vector lengths)
Cos(Fi) = (PA * PB) / (|PA| * |PB|)
Another solution could be using binary search on the vector N, whether P is close to C then the angle will be smaller and whether P is far from C then the angle will be bigger, being it suitable for a binary search.
how can i calculate the polynomial that has the tangent lines (1) y = x where x = 1, and (2) y = 1 where x = 365
I realize this may not be the proper forum but I figured somebody here could answer this in jiffy.
Also, I am not looking for an algorithm to answer this. I'd just like like to see the process.
Thanks.
I guess I should have mentioned that i'm writing an algorithm for scaling the y-axis of flotr graph
The specification of the curve can be expressed as four constraints:
y(1) = 1, y'(1) = 1 => tangent is (y=x) when x=1
y(365) = 1, y'(365) = 0 => tangent is (y=1) when x=365
We therefore need a family of curves with at least four degrees of freedom to match these constraints; the simplest type of polynomial is a cubic,
y = a*x^3 + b*x^2 + c*x + d
y' = 3*a*x^2 + 2*b*x + c
and the constraints give the following equations for the parameters:
a + b + c + d = 1
3*a + 2*b + c = 1
48627125*a + 133225*b + 365*c + d = 1
399675*a + 730*b + c = 0
I'm too old and too lazy to solve these myself, so I googled a linear equation solver to give the answer:
a = 1/132496, b = -731/132496, c = 133955/132496, d = -729/132496
I will post this type of question in mathoverflow.net next time. thanks
my solution in javascript was to adapt the equation of a circle:
var radius = Math.pow((2*Math.pow(365, 2)), 1/2);
var t = 365; //offset
this.tMax = (Math.pow(Math.pow(r, 2) - Math.pow(x, 2), 1/2) - t) * (t / (r - t)) + 1;
the above equation has the above specified asymptotes. it is part of a step polynomial for scaling an axis for a flotr graph.
well, you are missing data (you need another point to determine the polynomial)
a*(x-1)^2+b*(x-1)+c=y-1
a*(x-365)^2+b*(x-365)+c=y-1
you can solve the exact answer for b
but A depends on C (or vv)
and your question is off topic anyways, and you need to revise your algebra