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I took algebra a while ago and went as far as lines/planes/vector/points and products. It's all fuzzy to me now.

I want to solve the following problem:

Given 3 points that define a plane in space, I want to find the projection of a point Q on to said plane, said projection (point P) is defined by a light source (vector v).
Depending on the vector, there may be no shadow (projection) on to the plane.

So far I have made following conclusions:
-For there to be a shadow point(projection point P) to exist, the point Q has to be part of the line that includes the light source v.

-Depending on the direction and position of the light source v related to the plane, there could be no shadow even if the point Q is part of the line that vector v passes through.

To visualize your problem, let me simplify things; we can treat the more general case once we establish what's going on.

Let the plane be the familiar horizontal x-y plane from high school. Let the point Q lie on the vertical z axis at z0. It sounds like you are envisioning the light source to be like a laser, in other words, a pencil-like beam that looks like a line in 3-D (as opposed to an omni-directional light source). In this case, a shadow is produced only if the beam hits the point Q, as you have concluded.

To simplify even further, assume the light beam starts from somewhere above the x axis and travels down through the (transparent for now) point Q, lying in the x-z plane, at some angle A from the z axis. It will strike along the x axis at a distance x0 = z0・tan(A). This is where the shadow, P, of Q will be.

Now if the light beam is aligned parallel to the x axis then of course there will be no projection along he x-axis and no shadow, i.e., no point P. There will obviously also be no shadow if the light beam is pointed up, away from the x axis.

If this scenario is what you are interested in, then we can go to the general case. A plane in 3-D can be described by the equation (grab a textbook)

(1)     n1(x-x0) + n2(y-y0) + n3(z-z0) = 0

where P0 = (x0, y0, z0) = some point in the plane (arbitrary but given) and

<n1, n2, n3> = direction vector for a line perpendicular to the plane (the "n" can be thought of as standing for "normal");

the n's are the slopes of the perpendicular line, N, in the x, y and z directions. This definition comes from the fact that all points in the plane are perpendicular to the line N (i.e., the dot product of the vector from P0 to a point x,y,z with N is zero).

Now define the point Q = (xQ,yQ,zQ). The location and direction of the light source (beam) passing through Q will also define a line with direction parameters <a,b,c>. These are assumed known along with <n1,n2,n3>, (x0,y0,z0) and (xQ,yQ,zQ). A parametric eqn. for this line is

x = xQ + at
y = yQ + bt    (2)
z = zQ + ct

where t is just a parameter (real number). The light beam line is described in 3-D by eqns. (2) as s takes on values -∞ < t < ∞.

So now, the location of the shadow on the given plane, Ps = (xs, ys, zs), needs to be solved for. Setting x = xs, y = ys and z = zs in eqns. (2) and then substituting these into eqn. (1) gives a value for t

t = { n1・(x0-xQ) + n2・(y0-yQ) + n3・(z0-zQ) } / { n1 + n2 + n3 }

which is just a number based on known (given) quantities. This value of t gives a point along the light beam where it intersects with the plane (i.e., eqns. (2))

xs = xQ + at
yz = yQ + bt
zs = zQ + ct.

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randy patton


college mathematics, applied math, advanced calculus, complex analysis, linear and abstract algebra, probability theory, signal processing, undergraduate physics, physical oceanography


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J. Physical Oceanography, 1984 "A Numerical Model for Low-Frequency Equatorial Dynamics", with M. Cane

M.S. MIT Physical Oceanography, B.S. UC Berkeley Applied Math

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