Find the intersection of a line with a plane: Difference between revisions

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=={{header|APL}}==
=={{header|APL}}==
<lang APL>⍝ Find Intersection of a line with a plane
<lang APL>⍝ Find the intersection of a line with a plane
⍝ The intersection I belongs to a line defined by point L and vector V, means that t exists, so that I = L + tV
⍝ The intersection I belongs to a line defined by point L and vector V, translates to:
⍝ A real parameter t exists, that satisfies I = L + tV
⍝ I belongs to the plan defined by point P and normal vector N. This means that the vector IP is normal to vector N
⍝ I belongs to the plan defined by point P and normal vector N. This means that any two points of the plane make a vector normal ⍝ to vector N, I and P belong to the plane, so the vector IP is normal to N.
⍝ This translates to their scalar product being zero.
⍝ This translates to: The scalar product IP.N = 0.
⍝ (P - I).N = 0 <=> (P - L - tV).N = 0
⍝ (P - I).N = 0 <=> (P - L - tV).N = 0
⍝ Using distributivity, then associativity, the following equations are established:
⍝ Using distributivity, then associativity, the following equations are established:
⍝ (P - L - tV).N = (P - L).N - (tV).N = (P - L).N - t(V.N) = 0
⍝ (P - L - tV).N = (P - L).N - (tV).N = (P - L).N - t(V.N) = 0
⍝ so that: t = ((P - L).N) ÷ (V.N)
⍝ Which allows to resolve t: t = ((P - L).N) ÷ (V.N)
⍝ In APL, A.B is coded +/A x B
⍝ In APL, A.B is coded +/A x B
V ← 0 ¯1 ¯1
V ← 0 ¯1 ¯1
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N ← 0 0 1
N ← 0 0 1
P ← 0 0 5
P ← 0 0 5
dot ← {+/⍺×⍵}
dot ← { +/ ⍺ × ⍵ }
t ← ((P - L) dot N) ÷ V dot N
t ← ((P - L) dot N) ÷ V dot N
I ← L + t × V
I ← L + t × V
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