How to find Third Z with 3D Line with 2 3D Points and 1 Known 2D Point-CodePudding

I am trying to solve this, but not lucky:

$p1 = [0,0,0]; // x,y,z
$p2 = [20,10,1]; // x,y,z
$p3 = [30,15]; // x,y

$p3z = calc_z3($p1,$p2,$p3);

How to calculate the Z of P3?

CodePudding user response：

I see it like this:

l12 = sqrt( (x2-x1)^2   (y2-y1)^2 )
l13 = sqrt( (x3-x1)^2   (y3-y1)^2 )
z3 = z1 ((z2-z1)*l13/l12)

so:

                  | (x3-x1)^2   (y3-y1)^2 |^0.5
z3 = z1   (z2-z1)*|-----------------------|
                  | (x2-x1)^2   (y2-y1)^2 |

CodePudding user response：

I found this solution:

function calc_y3 ($x1,$y1,$x2,$y2,$x3) {
    $m = ($y2 - $y1) / ($x2 - $x1);
    $b = $y1 - $m * $x1;
    $y3 = $m * $x3   $b;
    return $y3;
}

function calc_z3 ($x1,$y1,$z1,$x2,$y2,$z2,$x3,$y3) {
    $x = $x2-$x1;
    $y = $y2-$y1;
    $z = $z2-$z1;
    if ($x == 0 and $y == 0) // it's a point
        return $z1;
    if ($x == 0) // it's a 2d line
        return calc_y3($y1,$z1,$y2,$z2,$y3);
    if ($y == 0) // it's a 2d line
        return calc_y3($x1,$z1,$x2,$z2,$x3);
    $t = ($y3-$y1)/$y;
    $z3 = $z1 $t*$z;
    return $z3;
}