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Hi, I am going to change your points $A, B, C, D$ and $E$ to $A(0,1,2), B(1,0,1), C(2,1,1), D(0,1/2,0)$ and $E(1,1,2)$ to make the problem slightly more interesting. With these coordinated a vector in the direction from $D$ to $E$ is $(1, 1/2, 2)$. Thus the equation of the line containing $D$ and $E$ is \[x = t, y = \frac12 + \frac12 t, z = 2t\] Notice that a point on the line is on the line segment from $D$ to $E$ if and only if $0 \leq t \leq 1.$ Now I need the equation of the plane containing the triangle $ABC.$ The vector from $A$ to $B$ is $\vec{u} = (1,-1,-1)$ and the vector from $A$ to $C$ is $\vec{v} = (2,1,0).$ The cross product of $\vec{u}$ and $\vec{v}$ is a vector $\vec{n} = (1,1,2)$ which is normal to the plane containing $A, B,$ and $C.$ Hence the equation of the plane containing $A,B$ and $C$ is \[1(x-0) +1(y-1) + 2(z-2) = 0 \mbox{ or } x +y +2z = 5.\] To determine the point $P$ where the line meets the plane substitute the equation of the line into the equation of the plane and solve for $t.$ I got $t = \large \frac{9}{10}$ and the point $P$ has coordinates $(\large \frac{18}{20}, \frac{19}{20}, \frac{36}{20}\normalsize ).$ Since $t$ satisfies $0 \leq t \leq 1$ the line segment intersect the plane containing the triangle. Finally let's determine if $P$ is in the triangle. To accomplish this I am going to use the vectors $\vec{u}, \vec{v}$ and $\vec{n}$ again. Use these three vectors for a base and write the point $P$ in this basis. That is find $p, q$ and $r$ so that \[p \vec{u} + q \vec{v} + r \vec{n} = P.\] The point $P$ is inside or on the boundary of the triangle if $p,q \geq 0$ and $p+q \leq1.$ (See Wolfram Math World.) The linear system \[p \vec{u} + q \vec{v} + r \vec{n} = P.\] is \begin{eqnarray*} Solving this system yields \[p = \frac{1}{10} \mbox{ and } q = \frac{-7}{40}.\] Since $q < 0$ the point $P$ is not inside or on the perimeter of the triangle. Write back if you need more help, RJ wrote back
The code you sent didn't arrive in a readable form and it doesn't matter since I am not a coder. I do have a suggestion however. On the stackoverflow page you use a technique in two dimensions involving the areas of four triangles to decide if $P$ is inside the triangle $ABC.$ You can use a similar technique in three dimensions if you know the coordinates of $A, B, C$ and $P,$ and $P$ is in the plane containing triangle $ABC.$ First use the formula for the distance $d$ between two points $(x_0, y_0, z_0)$ and $(x_1, y_1, z_1)$ in 3-space \[d = \sqrt{(x_1 - x_0)^2 + (y_1 - y_0)^2 + (z_1 - z_0)^2)}\] to find the lengths of $AB, BC, CA, AP, BP$ and $CP.$ Then use Heron's Formula to calculate the areas triangles $ABC, APB, APC,$ and $BPC$ and check to see if the area of triangle $ABC$ is the sum of the areas of the other three. The challenge here is to determine if $P$ is on the plane containing the triangle and if so to determine the coordinates of $P.$ The technique I described earlier does work. The question you originally asked is a geometric problem in three space and you are going to have to use some three dimensional geometry to solve it. Harley |
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